# Multiple, Sub-multiple Angles, and Transformation of Trigonometric Formula

Without using the calculator or table, find the value of:
sin100°.sin120°.sin140°.sin160°

Here,
= sin100°.sin120°.sin140°.sin160°
= sin(90° + 10°).sin(90° + 30°).sin(90° + 50°).sin(90° + 70°)
= cos10°.cos30°.cos50°.cos70°
= cos10°.$\frac{\sqrt{3}}{2}$.$\frac{2}{2}$.cos50°.cos70°
= $\frac{\sqrt{3}}{4}$.cos10°.(2cos50°.cos70°)
= $\frac{\sqrt{3}}{4}$.cos10°.{cos(50° + 70°) + cos(50° - 70°)}
= $\frac{\sqrt{3}}{4}$.cos10°{cos120° + cos(-20°)}
= $\frac{\sqrt{3}}{4}$.$\frac{2}{2}$.(cos10°.cos120° + cos10°.cos20°) [cos(-θ) = cos(θ)]
= $\frac{\sqrt{3}}{8}$(2cos10°.cos120° + 2cos10°.cos20°)
= $\frac{\sqrt{3}}{8}${2cos10°.-$\frac{1}{2}$ + cos(10° + 20°) + cos(10° - 20°)}
= $\frac{\sqrt{3}}{8}$(-cos10° + cos30° + cos10°) [cos(-θ) = cos(θ)]
= $\frac{\sqrt{3}}{8}$ × $\frac{\sqrt{3}}{2}$
= $\frac{3}{16}$

$\frac{\text{sec4θ - 1}}{sec2θ - 1}$ = tan4θ.cotθ

Here,
$\frac{\text{sec4θ - 1}}{sec2θ - 1}$ = tan4θ.cotθ

Taking L.H.S,
= $\frac{\text{sec4θ - 1}}{sec2θ - 1}$
= $\frac{\text{1 - cos4θ}}{cos4θ}$ × $\frac{cos2θ}{\text{1 - cos2θ}}$
= $\frac{\text{1 - cos2.2θ}}{cos4θ}$ × $\frac{cos2θ}{\text{1 - }(2cos^{2}θ \text{ - 1})}$
= $\frac{\text{1 - (}2cos^{2}2θ\text{ - 1)}}{cos4θ}$ × $\frac{cos2θ}{\text{1 - }2cos^{2}θ \text{ + 1}}$
= $\frac{\text{2(1 - }cos^{2}2θ)}{cos4θ}$ × $\frac{cos2θ}{\text{2(1 - }cos^{2}θ)}$
= $\frac{sin^{2}2θ}{cos4θ}$ × $\frac{cos2θ}{sin^2θ}$
= $\frac{sin2θ \times 2sinθcosθ}{cos4θ}$ × $\frac{cos2θ}{sin^{2}θ}$
= $\frac{2sin2θcos2θ}{cos4θ}$ × $\frac{sinθcosθ}{sin^{2}θ}$
= $\frac{sin4θ}{cos4θ}$ × cotθ

∴ tan4θ.cotθ = R.H.S

8(sin6p + cos6p) = 5 + 3cos4p

Here,
8(sin6p + cos6p) = 5 + 3cos4p

Taking L.H.S,
= 8 {(sin2p)3 + (cos2p)3}
= 8 (sin2p + cos2p) (sin4p - sin2pcos2p + cos4p)
= 8 {(sin2p + cos2p)2 - 2sin2pcos2p - sin2pcos2p}
= 8 {1 - 3sin2pcos2p}
= 8 {1 - $\frac 34$(2sinpcosp)2}
= 8 {1 - $\frac 34$(sin2p)2}
= 8 {1 - $\frac 38$(2sin22p)}
= $\frac 88${8 - 3(1 - cos2.2p)}
= 8 -3 + 3cos4p

∴ 5 + 3cos4p = R.H.S

Prove that: $\frac{\text{1 - cosα + sinα}}{\text{1 + cosα + sinα}}$ =tan$\frac{α}{2}$

Here,
$\frac{\text{1 - cosα + sinα}}{\text{1 + cosα + sinα}}$ =tan$\frac{α}{2}$

Taking L.H.S,
= $\frac{\text{1 - cosα + sinα}}{\text{1 + cosα + sinα}}$
= $\frac{\text{1 - }cos2.\fracα2\text{ + sinα}}{\text{1 + }cos2.\fracα2\text{ + sinα}}$
= $\frac{2sin^2\frac α2 \text{ + } sin2.\frac α2}{2cos^2\frac α2 \text{ + } sin2.\fracα2}$
= $\frac{2sin^2\frac α2 \text{ + } 2sin\frac α2 cos\frac α2}{2cos^2\frac α2 \text{ + } 2sin\frac α2 cos\frac α2}$
= $\frac{2sin\frac α2(sin\frac α2 \text{ + }cos\frac α2)}{2cos\frac α2(cos\frac α2 \text{ + }sin\frac α2)}$

∴ tan$\frac α2$ = R.H.S

tanθ + 2tan2θ + 4cot4θ = cotθ

Here,
tanθ + 2tan2θ + 4cot4θ = cotθ

Taking L.H.S,
= tanθ + 2tan2θ + 4cot4θ
= tanθ + 2tan2θ + $\frac{4cos4θ}{sin4θ}$
= tanθ + 2tan2θ + $\frac{4cos4θ}{2sin2θ . cos2θ}$
= tanθ + 2tan2θ + $\frac{2cos4θ}{2sinθ . cosθ . cos2θ}$
= tanθ + $\frac{2sin2θ}{cos2θ}$ + $\frac{cos4θ}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{\text{2sin2θ . sinθ . cosθ + cos4θ}}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{\text{sin2θ . sin2θ + cos2.2θ}}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{sin^{2}2θ\text{ + 1 - }2sin^{2}2θ}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{\text{1 - }sin^{2}2θ}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{cos^{2}2θ}{sinθ . cosθ . cos2θ}$
= tanθ + $\frac{cos2θ}{sinθ . cosθ}$ × $\frac 22$
= tanθ + $\frac{2cos2θ}{sin2θ}$

= tanθ + 2cot2θ
= $\frac{sinθ}{cosθ}$ + $\frac{2cos2θ}{2sinθ.cosθ}$
= $\frac{sin^2θ \text{ + cos2θ}}{sinθcosθ}$
= $\frac{sin^2θ \text{ + 1 - }2sin^{2}θ}{sinθcosθ}$
= $\frac{\text{1 - }sin^{2}θ}{sinθcosθ}$
= $\frac{cos^{2}θ}{sinθcosθ}$

= cotθ

sin20°.sin30°.sin40°.sin80° = $\frac{\sqrt3}{16}$

Here,
sin20°.sin30°.sin40°.sin80° = $\frac{\sqrt3}{16}$

Taking L.H.S,
= sin20°.sin30°.sin40°.sin80°
= $\frac{sin20°}{2}$ × $\frac 22$sin40°.sin80°
= $\frac{sin20°}{4}${cos(40° - 80°) - cos(40° + 80°)}
= $\frac{sin20°}{4}${cos(-40°) - cos120°}
= $\frac{sin20°}{4}${cos40° + $\frac 12$} [cos(-θ) = cosθ]
= $\frac{sin20°}{8}$(2cos40° + 1)
= $\frac 18$(2sin20°.cos40° + sin20°)
= $\frac 18${sin(20° + 40°) + sin(20° - 40°) + sin20°}
= $\frac 18$(sin60° - sin20° + sin20°)

$\frac{\sqrt3}{16}$ = R.H.S

cos3A.cos3A + sin3A.sin3A = cos32A

Here,
cos3A.cos3A + sin3A.sin3A = cos32A

Taking L.H.S,
= cos3A.cos3A + sin3A.sin3A
= $\frac{4cos^{3}A.cos3A \text{ + }4sin^{3}A.sin3A}{4}$
= $\frac 14$ {(cos3A + 3cosA)cos3A + (3sinA - sin3A)sin3A}
= $\frac 14$ {cos23A + 3cosA.cos3A + 3sinA.sin3A - sin23A}
= $\frac 14$ {cos2.3A + 3(cosA.cos3A + sinA.sin3A)}
= $\frac 14$ {cos6A + 3cos(-2A)}
= $\frac 14$ {cos6A + 3cos2A} [cos(-θ) = cosθ]
= $\frac 14$ {cos6A + 4cos32A - cos3.2A}
= $\frac 14$ (cos6A + 4cos32A - cos6A)

∴ cos32A = R.H.S

cos2A + sin2A.cos2B = cos2B + sin2B.cos2A

Here,
cos2A + sin2A.cos2B = cos2B + sin2B.cos2A

Taking L.H.S,
= cos2A + sin2A(1 - 2sin2B)
= cos2A + sin2A - 2sin2A.sin2B
= 1 - 2sin2A.sin2B
= cos2B + sin2B - 2sin2A.sin2B
= cos2B + sin2B(1 - 2sin2A)

∴ cos2B + sin2B.cos2A = R.H.S

If $\frac{1}{sinA}$ + $\frac{1}{cosA}$ = $\frac{1}{sinB}$ + $\frac{1}{cosB}$, prove that: cot$\left(\frac{\text{A + B}}{2}\right)$ = tanA.tanB

Given: $\frac{1}{sinA}$ + $\frac{1}{cosA}$ = $\frac{1}{sinB}$ + $\frac{1}{cosB}$
To prove: cot$\left(\frac{\text{A + B}}{2}\right)$ = tanA.tanB

Now, $\frac{1}{sinA}$ - $\frac{1}{cosB}$ = $\frac{1}{sinB}$ - $\frac{1}{cosA}$ or, $\frac{\text{cosB - sinA}}{sinA.cosB}$ = $\frac{\text{cosA - sinB}}{sinB.cosA}$
or, cosA.cosB.sinB - sinA.sinB.cosA = sinA.cosA.cosB - sinA.sinB.cosB
Dividing both sides by cosA.cosB.cosC, we get,
or, $\frac{sinB}{cosC}$ - $\frac{tanA.tanB.cosA}{cosC}$ = $\frac{tanA.cosA}{cosC}$ - $\frac{tanA.tanB.cosB}{cosC}$
or, sinB - tanA.tanB.cosA = tanA.cosA - tanA.tanB.cosB
or, sinB - tanA.cosA = tanA.tanB(cosA - cosB)
or, sinB - $\frac{sinA.cosA}{cosA}$ = tanA.tanB(cosA - cosB)
or, $\frac{\text{sinB - sinA}}{\text{cosA - cosB}}$ = tanA.tanB

or, $\frac{2sin\left(\frac{\text{B - A}}{2}\right)cos\left(\frac{\text{B + A}}{2}\right)}{2sin\left(\frac{\text{B - A}}{2}\right)sin\left(\frac{\text{B + A}}{2}\right)}$ = tanA.tanB
or, cot$\left(\frac{\text{B + A}}{2}\right)$ = tanA.tanB
∴ cot$\left(\frac{\text{A + B}}{2}\right)$ = tanA.tanB

8$\left(\text{1 + }sin\frac{\pi}{8}\right)\left(\text{1 + }sin\frac{3\pi}{8}\right)\left(\text{1 - }sin\frac{5\pi}{8}\right)\left(\text{1 - }sin\frac{7\pi}{8}\right)$ = 1

Here, 8$\left(\text{1 + }sin\frac{\pi}{8}\right)\left(\text{1 + }sin\frac{3\pi}{8}\right)\left(\text{1 - }sin\frac{5\pi}{8}\right)\left(\text{1 - }sin\frac{7\pi}{8}\right)$ = 1

Taking L.H.S, = 8$\left(\text{1 + }sin\frac{\pi}{8}\right)\left(\text{1 + }sin\frac{3\pi}{8}\right)\left(\text{1 - }sin\frac{5\pi}{8}\right)\left(\text{1 - }sin\frac{7\pi}{8}\right)$
= 8$\left(\text{1 + }sin\frac{\pi}{8}\right)\{(\text{1 - }sin\left(\pi\text{ - }\frac{\pi}{8}\right)\}\left(\text{1 + }sin\frac{3\pi}{8}\right)\{(\text{1 - }sin\left(\pi\text{ - }\frac{3\pi}{8}\right)\}$
= 8$\left(\text{1 + }sin\frac{\pi}{8}\right)\left(\text{1 - }sin\frac{\pi}{8}\right)\left(\text{1 + }sin\frac{3\pi}{8}\right)\left(\text{1 - }sin\frac{3\pi}{8}\right)$
= 8$\left(\text{1 - }sin^{2}\frac{\pi}{8}\right)\left(\text{1 - }sin^{2}\frac{3\pi}{8}\right)$
= 8$\left(\text{1 - }sin^{2}\frac{3\pi}{8}\text{ - }sin^{2}\frac{\pi}{8}\text{ + }sin^{2}\frac{3\pi}{8}sin^{2}\frac{\pi}{8}\right)$
= 8$\{\text{1 - }cos^{2}\left(\frac{\pi}{2}\text{ - }\frac{\pi}{8}\right)\text{ - }sin^{2}\frac{\pi}{8}\text{ + }cos^{2}\left(\frac{\pi}{2}\text{ - }\frac{\pi}{8}\right)sin^{2}\frac{\pi}{8}\}$
= 8$\{1 \text{ - } \left(cos^{2}\frac \pi8 \text{ + }sin^{2}\frac \pi8\right)\text{ + }cos^{2}\frac{\pi}{8}sin^{2}\frac \pi8\}$
= 8$\{\text{1 - 1 + }\frac 14\left(2sin\frac{\pi}{8}cos\frac \pi8\right)^2\}$
= $\frac 84\left(sin2 . \frac \pi8\right)^2$
= 2$\left(sin\frac \pi4\right)^2$
= 2$\left(\frac{1}{\sqrt2}\right)^2$
= $\frac 22$
= 1

8cos10°.cos50°.cos70° = $\sqrt3$

Here,
8cos10°.cos50°.cos70° = $\sqrt3$

Taking L.H.S,
= 4cos10°.2cos50°cos70°
= 4cos10°{cos(50° + 70°) + cos(50° - 70°)}
= 4cos10°(cos120° + cos20°) [cos(-θ) = cosθ]
= 4cos10°(-$\frac{1}{2}$ + cos20°)
= 2(-cos10° + 2cos10°cos20°)
= 2{-cos10° + cos(10° + 20°) + cos(10° - 20°)}
= 2{-cos10° + cos30° + cos(-10)°}
= 2(-cos10° + $\frac{\sqrt3}{2}$ + cos10°)
= $\sqrt3$

4cosA.cos(60° - A).cos(60° + A) = cos3A

Here,
4cosA.cos(60° - A).cos(60° + A) = cos3A

Taking L.H.S,
= 2cosA.2cos(60° - A).cos(60° + A)
= 2cosA{cos(60° - A + 60° + A) + cos(60° - A - 60° - A)}
= 2cosA{cos120° + cos(-2A)}
= 2(-$\frac{cosA}{2}$ + cosA.cos2A) [cos(-θ) = cosθ]
= -cosA + 2xosA.cos2A
= -cosA + cos(A + 2A) + cos(A - 2A)
= -cosA + cos3A + cosA

∴ cos3A = R.H.S

16sin20°.sin40°.sin60°.sin80° = 3

Here,
16sin20°.sin40°.sin60°.sin80° = 3

Taking L.H.S,
= 16sin20°.sin40°.sin60°.sin80°
= 16sin20°.sin40°.$\frac{\sqrt3}{2}$.sin80°
= 4$\sqrt3$.sin20°.2sin80°sin40°
= 4$\sqrt3$.sin20°{cos(80° - 40°) - cos(80° + 40°)}
= 4$\sqrt3$.sin20°(cos40° - cos120°)
= 4$\sqrt3$.sin20°(cos40° + $\frac 12$)
= 2$\sqrt3$.sin20°(2cos40° + 1)
= 2$\sqrt3$.(2sin20°.cos40° + sin20°)
= 2$\sqrt3$.{sin(20° + 40°) + sin(20° - 40°) + sin20°}
= 2$\sqrt3$.{sin60° - sin20° + sin20°}
= 2$\sqrt3.\frac{\sqrt3}{2}$
= 3

If x + y = 45°, prove that: $\frac{\text{(cotx + 1)(coty + 1)}}{cotx.coty}$ = 2

Given: x + y = 45°
To prove: $\frac{\text{(cotx + 1)(coty + 1)}}{cotx.coty}$ = 2

Now, x + y = 45° Taking 'cot' ratio on both side, we get,
or, cot(x + y) = cot45°
or, $\frac{\text{cotx.coty - 1}}{cotx + coty}$ = 1
or, cotx.coty = cotx + coty + 1
Adding 'cotx.coty' on both side, we get,
or, 2cotx.coty = cotx + coty + 1 + cotx.coty
or, 2cotx.coty = 1(cotx + 1) + coty(1 + cotx)
or, 2cotx.coty = (cotx + 1)(1 + coty)
$\frac{\text{(cotx + 1)(coty + 1)}}{cotx.coty}$ = 2

$\frac{sin^{2}α\text{ - }sin^{2}β}{\text{sinα.cosα - sinβ.cosβ}}$ = tan(α + β)

Here,
$\frac{sin^{2}α\text{ - }sin^{2}β}{\text{sinα.cosα - sinβ.cosβ}}$ = tan(α + β)

Taking R.H.S,
= tan(α + β)
= $\frac{\text{sin(α + β)}}{\text{cos(α + β)}}$
= $\frac{\text{sinαcosβ + cosαsinβ}}{\text{cosαcosβ - sinαsinβ}}$ × $\frac{\text{sinαcosβ - cosαsinβ}}{\text{sinαcosβ - cosαsinβ}}$ = $\frac{sin^{2}αcos^{2}β \text{ - }cos^{2}αsin^{2}β}{sinα.cosα.cos^{2}β \text{ - }cos^{2}α.sinβ.cosβ \text{ - }sin^{2}α.sinβ.cosβ \text{ + }sinα.cosα.sin^{2}β}$
= $\frac{sin^{2}α(\text{1 - }sin^{2}β) \text{ - }(\text{1 - }sin^{2}α)sin^{2}β}{sinα.cosα(cos^{2}β \text{ + }sin^{2}β) \text{ - } sinβ.cosβ(cos^{2}α \text{ + } sin^{2}α)}$
= $\frac{sin^{2}α \text{ - }sin^{2}αsin^{2}β \text{ - }sin^{2}β \text{ + } sin^{2}αsin^{2}β}{\text{sinα.cosα.1 - sinβ.cosβ.1}}$

$\frac{sin^{2}α\text{ - }sin^{2}β}{\text{sinα.cosα - sinβ.cosβ}}$ = L.H.S

sinθ.sin(60° - θ).sin(60° + θ) = $\frac 14$sin3θ

Here,
sinθ.sin(60° - θ).sin(60° + θ) = $\frac 14$sin3θ

Taking L.H.S,
= sinθ.sin(60° - θ).sin(60° + θ)
= $\frac 12$sinθ.2sin(60° - θ).sin(60° + θ)
= $\frac 12$sinθ {cos(60° - θ - 60° - θ) - cos(60° - θ + 60° - θ)}
= $\frac 12$sinθ {cos(-2θ) - cos120°}
= $\frac 12$sinθ {cos2θ + $\frac 12$}
= $\frac 14$ (2sinθ.cos2θ + sinθ)
= $\frac 14$ {sin(θ + 2θ) + sin(θ - 2θ) + sinθ}
= $\frac 14$ (sin3θ - sinθ + sinθ)

$\frac 14$sin3θ = R.H.S

$\sqrt3$ cosec20° - sec20° = 4

Here,
$\sqrt3$ cosec20° - sec20° = 4

Taking L.H.S,
= $\sqrt3$ cosec20° - sec20°
= $\frac{\sqrt3}{sin20°}$ - $\frac{1}{cos20°}$
= $\frac{\sqrt3 cos20° \text{ - sin20°}}{sin20°.cos20°}$
= $\frac{cot30°.cos20° \text{ - sin20°}}{sin20°.cos20°}$
= $\frac{cos30°.cos20° \text{ - sin30°sin20°}}{sin30°.sin20°.cos20°}$
= $\frac{2 \times \text{cos(30° + 20°)}}{sin30°.2sin20°.cos20°}$
= $\frac{2cos50°}{sin30°.sin40°}$
= $\frac{\text{2cos(90° - 40°)}}{\frac 12 sin40°}$

= $\frac{4sin40°}{sin40°}$

∴ 4 = R.H.S

sin4x = $\frac 18$ (3 - 4cos2x + cos4x)

Here,
sin4x = $\frac 18$ (3 - 4cos2x + cos4x)

Taking L.H.S,
= $\frac 18$ (3 - 4cos2x + cos4x)
= $\frac 18$ {3 - 4(1 - 2sin2x) + 1 - 2 sin22x}
= $\frac 18$ (3 - 4 + 8sin2x + 1 - 2 sin22x)
= $\frac 18$ (-1 + 8sin2x + 1 - 2sin22x)
= $\frac 28$ (4sin2x - sin22x)
= $\frac 14$ {4sin2x - (2sinx.cosx)2}
= $\frac 14$ (4sin2x - 4sin2x.cos2x)
= $\frac 44$ sin2x (1 - cos2x)
= sin2x.sin2x

∴ sin4x = R.H.S

If 2tanα = 3tanβ, prove that: tan(α - β) = $\frac{sin2β}{\text{5 - cos2β}}$

Given: 2tanα = 3tanβ
tanα = $\frac{3tanβ}{2}$ ---- (A)
To prove: tan(α - β) = $\frac{sin2β}{\text{5 - cos2β}}$

Taking L.H.S,
= tan(α - β)
= $\frac{\text{tanα - tanβ}}{\text{1 + tanα.tanβ}}$
= $\frac{\frac{3tanβ}{2} \text{ - tanβ}}{\text{1 + }\frac{3tanβ}{2}tanβ}$ [From (A)]
= $\frac{\text{3tanβ - 2tanβ}}{2}$ × $\frac{2}{\text{2 + }3tan^{2}β}$
= $\frac{tanβ}{\text{2 + }3tan^{2}β}$
= $\frac{\frac{sinβ}{cosβ}}{\text{2 + }\frac{3sin^{2}β}{cos^{2}β}}$
= $\frac{sinβ}{cosβ}$ × $\frac{cos^2{β}}{2cos^{2}β \text{ + }3sin^{2}β}$
= $\frac{sinβ.cosβ}{2cos^{2}β \text{ + }3sin^{2}β}$
= $\frac{2sinβ.cosβ}{2.2cos^{2}β \text{ + }3.2sin^{2}β}$
= $\frac{2sinβ.cosβ}{\text{2(1 + cos2β) + 3(1 - cos2β)}}$
= $\frac{sin2β}{\text{2 + 2cos2β + 3 - 3cos2β}}$

$\frac{sin2β}{\text{5 - cos2β}}$ = R.H.S