# Multiple Angles

Prove that: sin4B + cos4B = 1 - $\frac 12$sin22B

Here,
sin4B + cos4B = 1 - $\frac 12$sin22B

Taking L.H.S,
= sin4B + cos4B
= (sin2B)2 + (cos2B)2
= (sin2B + cos2B)2 - 2sin2B.cos2B
= 1 - $\frac 12$(2sinB.cosB)2
= 1 - $\frac 12$(sin2B)2
= 1 - $\frac 12$sin22B

Prove that: cos4A + sin4A = 1 - $\frac 12$sin22A

Here,
cos4A + sin4A = 1 - $\frac 12$sin22A

Taking L.H.S,
= cos4A + sin4A
= (cos2A)2 + (sin2A)2
= (cos2A + sin2A)2 - 2sin2A.cos2A
= 1 - $\frac 12$(2sinA.cosA)2
= 1 - $\frac 12$(sin2A)2
= 1 - $\frac 12$sin22A

Prove that: $\frac{\text{sinθ + sin2θ}}{\text{1 + cosθ + cos2θ}}$ = tanθ

Here,
$\frac{\text{sinθ + sin2θ}}{\text{1 + cosθ + cos2θ}}$ = tanθ

Taking L.H.S,
= $\frac{\text{sinθ + sin2θ}}{\text{1 + cosθ + cos2θ}}$
= $\frac{\text{sinθ + 2sinθcosθ}}{\text{1 + cosθ + }2cos^2θ\text{ - 1}}$
= $\frac{\text{sinθ (1 + 2cosθ)}}{cosθ (1 + 2cosθ)}$
∴ tanθ = R.H.S

If sinA = $\frac{1}{2}$, find the value of sin3A.

Here,
sinA = $\frac{1}{2}$

Now,
sin3A = 3sinA - 4sin3A = $\frac{3}{2}$ - 4$\left(\frac{1}{2}\right)^3$
= $\frac{3}{2}$ - $\frac{4}{8}$
= $\frac{3}{2}$ - $\frac{1}{2}$
= 1

If sinθ = $\frac{3}{5}$, find the value of cos2θ.

Here,
sinθ = $\frac{3}{5}$

Now,
cos2θ = 1 - 2sin2θ = 1 - 2$\left(\frac{3}{5}\right)^2$
= 1 - $\frac{18}{25}$
= $\frac{7}{25}$

Prove that: cosec2θ - cot2θ = tanθ

Here, cosec2θ - cot2θ = tanθ

Taking L.H.S,
= cosec2θ - cot2θ
= $\frac{1}{sin2θ}$ - $\frac{cos2θ}{sin2θ}$
= $\frac{\text{1 - cos2θ}}{sin2θ}$
= $\frac{\text{1 - }(2cos^2θ \text{ - 1})}{sin2θ}$
= $\frac{\text{2 - }2cos^2θ}{sin2θ}$
= $\frac{\text{2(1 - }cos^2θ)}{2sinθcosθ}$
= $\frac{sinθ}{sinθcosθ}$
∴ tanθ = R.H.S

If cosθ = $\frac{4}{5}$, find the value of cos3θ.

Here,
cosθ = $\frac{4}{5}$

Now,
cos3θ = 4cos3θ - 3cosθ = 4$\left(\frac{4}{5}\right)^3$ - 3$\left(\frac{4}{5}\right)$
= 4$\left(\frac{64}{125}\right)$ - $\left(\frac{12}{5}\right)$
= -$\frac{44}{125}$

Prove that: $\frac{sin2A}{\text{1 + cos2A}}$ = tanA

Here,
$\frac{sin2A}{\text{1 + cos2A}}$ = tanA

Taking L.H.S,
= $\frac{sin2A}{\text{1 + cos2A}}$
= $\frac{2sinAcosA}{\text{1 + }2cos^2A \text{ - 1}}$
∴ tanA = R.H.S

If sinA = $\frac{3}{4}$ find the value of cos2A.

Here,
sinA = $\frac{3}{4}$

Now,
cos2A = 2cos2A - 1 = 2(1 - sin2A) - 1
= 2$\left(\text{1 - }\frac{9}{16}\right)$
= $\frac{\text{14 - 16}}{16}$
= -$\frac{1}{8}$

Prove that: sinA.cos2A = $\frac{1}{4}$sin4A.secA

Here,
sinA.cos2A = $\frac{1}{4}$sin4A.secA

Taking R.H.S,
= $\frac{1}{4}$sin4A.secA
= $\frac{sin2.2A}{4cosA}$
= $\frac{2sin2Acos2A}{4cosA}$
= $\frac{\cancel{2}sinA\cancel{cosA}cos2A}{\cancel{2cosA}}$
∴ sinA.cos2A = L.H.S

2sin2(45 - A) = 1 - sin2A.

Here,
2sin2(45 - A) = 1 - sin2A

Taking L.H.S,
= 2sin2(45 - A)
= 1 - cos2(45 - A)
= 1 - cos(90 - 2A)
∴ 1 - sin2A = R.H.S