# Sub-multiple Angles

If sin$\frac A3$ = $\frac 12 (a + \frac 1a)$, prove that sinA = $-\frac 12 (a^3 + \frac {1}{a^3})$

Given: sin$\frac A3$ = $\frac 12 (a + \frac 1a)$
To Prove: sinA = $-\frac 12 (a^3 + \frac {1}{a^3})$

Now,
sinA = sin3$\frac A3$ = 3sin$\frac A3$ - 4sin3$\frac A3$
= 3 × $\frac 12 (a + \frac 1a)$ - 4 ×$\{\frac 12 (a + \frac 1a)\}^3$
= $\frac 32 (a + \frac 1a)$ - $\frac 48 (a + \frac 1a)^3$
= $\frac 32 (a + \frac 1a)$ - $\frac 12 \{a^3 + \frac{1}{a^3} + 3.a.\frac 1a (a + \frac 1a) \}$
= $\frac 32 (a + \frac 1a)$ - $\frac 12 (a^3 + \frac{1}{a^3}) - \frac 32 (a + \frac 1a)$

∴ sinA = $-\frac 12 (a^3 + \frac {1}{a^3})$

If cos$\frac B3$ = $\frac 12 (m + \frac 1m)$, prove that cosB = $\frac 12 (m^3 + \frac {1}{m^3})$

Given: cos$\frac B3$ = $\frac 12 (m + \frac 1m)$
To prove: cosB = $\frac 12 (m^3 + \frac {1}{m^3})$

Now,
cosB = cos3$\frac B3$ = 4cos3$\frac B3$ - 3cos$\frac B3$
= 4 ×$\{\frac 12 (m + \frac 1m)\}^3$ - 3 × $\frac 12 (m + \frac 1m)$
= $\frac 48 (m + \frac 1m)^3$ - $\frac 32 (m + \frac 1m)$
= $\frac 12 \{m^3 + \frac{1}{m^3} + 3.a.\frac 1m (m + \frac 1m) \}$ - $\frac 32 (a + \frac 1m)$
= $\frac 12 (m^3 + \frac{1}{m^3}) + \frac 32 (m + \frac 1m)$ - $\frac 32 (m + \frac 1m)$

∴ cosB = $\frac 12 (m^3 + \frac {1}{m^3})$

If tan$\frac A2$ = $\frac{3}{4}$, find the value of sinA.

Given: tan$\frac A2$ = $\frac{3}{4}$

Now,
sinA = sin2.$\frac A2$ = $\frac{2tan\frac A2}{\text{1 + }tan^2\frac{A}{2}}$
= $\frac{2.\frac 34}{\text{1 +}\frac{9}{16}}$
= $\frac 64$ × $\frac{16}{25}$
= $\frac{24}{25}$

If sin$\frac A3$ = $\frac{4}{5}$, find the value of sinA.

Given: sin$\frac A3$ = $\frac{4}{5}$

Now,
sinA = sin3$\frac A3$ = 3sin$\frac A3$ - 4sin3$\frac A3$
= 3 × $\frac 45$ - 4 ×$\left(\frac 45\right)^3$
= $\frac{12}{5}$ - $\frac{256}{125}$
= $\frac{44}{125}$

Prove that: $\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}}$ =tan$\frac{θ}{2}$

Here,
$\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}}$ =tan$\frac{θ}{2}$

Taking L.H.S,
= $\frac{\text{1 - }cos2.\fracθ2\text{ + sinθ}}{\text{1 + }cos2.\fracθ2\text{ + sinθ}}$
= $\frac{2sin^2\frac θ2 \text{ + } sin2.\frac θ2}{2cos^2\frac θ2 \text{ + } sin2.\fracθ2}$
= $\frac{2sin^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}{2cos^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}$
= $\frac{2sin\frac θ2(sin\frac θ2 \text{ + }cos\frac θ2)}{2cos\frac θ2(cos\frac θ2 \text{ + }sin\frac θ2)}$
= $\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}}$

∴ tan$\frac θ2$ = R.H.S

Prove that: $\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}}$ = tan$\frac{θ}{2}$

Here,
$\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}}$ = tan$\frac{θ}{2}$

Taking L.H.S,
= $\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}}$ = $\frac{sin\frac{θ}{2}\text{ + }sin2.\frac θ2}{\text{1 + }cos2.\fracθ2\text{ + }cos\frac{θ}{2}}$
= $\frac{sin\frac{θ}{2}\text{ + }2sin.\frac θ2 cos\frac θ2}{2cos^2\fracθ2\text{ + }cos\frac{θ}{2}}$
= $\frac{sin\frac{θ}{2}\text{(1 + }2cos\frac θ2)}{cos\frac θ2(2cos\fracθ2\text{ + 1)}}$

∴ tan$\frac θ2$ = R.H.S

Prove that: cosecθ - cotθ = tan$\frac{θ}{2}$

Here,
cosecθ - cotθ = tan$\frac{θ}{2}$

Taking L.H.S,
= cosecθ - cotθ
= $\frac{1}{sinθ}$ - $\frac{cosθ}{sinθ}$
= $\frac{\text{1 - cosθ}}{sinθ}$
= $\frac{\text{1 - }cos2.\frac θ2}{sin2.\frac θ2}$
= $\frac{2sin^2\frac θ2}{2sin\fracθ2 cos\fracθ2}$

∴ tan$\frac θ2$ = R.H.S

Prove that: cot$\frac{A}{2}$ - tan$\frac{A}{2}$ = 2cotA

Here,
cot$\frac{A}{2}$ - tan$\frac{A}{2}$ = 2cotA

Taking L.H.S,
= $\frac{cos\frac A2}{sin\frac A2}$ - $\frac{sin\frac A2}{cos\frac A2}$
= $\frac{cos^2\frac A2 \text{ - }sin^2\frac A2}{sin\frac A2 cos\frac A2}$
= $\frac{cos2.\frac A2}{sin\frac A2 cos\frac A2}$ × $\frac 22$
= $\frac{2cosA}{sin2.\frac A2}$
= $\frac{2cosA}{sinA}$

∴ 2cotA = R.H.S

Prove that: $\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}}$ = cot$\frac{θ}{2}$

Here,
$\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}}$ = cot$\frac{θ}{2}$

Taking L.H.S,
= $\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}}$
= $\frac{\text{1 + }cos2.\fracθ2\text{ + sinθ}}{\text{1 - }cos2.\fracθ2\text{ + sinθ}}$
= $\frac{2cos^2\frac θ2 \text{ + } sin2.\frac θ2}{2sin^2\frac θ2 \text{ + } sin2.\fracθ2}$
= $\frac{2cos^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}{2sin^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}$
= $\frac{2cos\frac θ2(cos\frac θ2 \text{ + }sin\frac θ2)}{2sin\frac θ2(sin\frac θ2 \text{ + }cos\frac θ2)}$

∴ cot$\frac θ2$ = R.H.S

If cos$\frac{α}{2}$ = $\frac{1}{2}$, find the value of sinα.

Here,
cos$\frac{α}{2}$ = $\frac{1}{2}$

Now,
sinα = sin2.$\frac α2$ = 2sin$\frac α2$.cos$\frac α2$
= 2 × $\sqrt{\text{1 - }cos^2\frac α2}$ × cos$\frac α2$
= 2 × $\sqrt{\text{1 - }\frac 14}$ × $\frac 12$
= $\sqrt{\frac{4 - 1}{4}}$
= $\frac{\sqrt{3}}{2}$

If cos$\frac{α}{3}$ = $\frac{1}{2}$, find the value of sinα.

Here,
cos$\frac{α}{3}$ = $\frac{1}{2}$

Now,
sinα = sin3.$\frac α3$ = 3sin$\frac α3$ - 4sin3$\frac α3$
= 3$\sqrt{\text{1 - }cos^2\frac α3}$ - 4$\left(\sqrt{\text{1 - }cos^2\frac α3}\right)^3$
= 3$\sqrt{\text{1 - }\frac 14}$ - 4$\left(\sqrt{\text{1 - }\frac 14}\right)^3$
= 3 × $\sqrt{\frac 34}$ - 4 × $\sqrt{\frac 34} \sqrt{\frac 34} \sqrt{\frac 34}$
= $\frac{3\sqrt{3}}{2}$ - 4 × $\frac 34$ × $\frac{\sqrt3}{2}$
= 0

If cos45° = $\frac{1}{\sqrt{2}}$, prove that: sin$\left(22\dfrac 12\right)$° = $\frac{\sqrt{2 - \sqrt{2}}}{2}$

Given: cos45° = $\frac{1}{\sqrt{2}}$
To prove: sin$\frac{45}{2}^°$ = $\frac{\sqrt{2 - \sqrt{2}}}{2}$

Now,
cos45° = $\frac{1}{\sqrt{2}}$ or, cos2.$\frac{45}{2}$ = $\frac{1}{\sqrt2}$
or, 2cos2$\frac{45}{2}$ = $\frac{1}{\sqrt2}$ + 1
or, 2$\left(\text{1 - }sin^2\frac{45}{2}\right)$ = $\frac{\text{1 + }\sqrt2}{\sqrt2}$
or, $\text{1 - }sin^2\frac{45}{2}$ = $\frac{\text{1 + }\sqrt2}{2\sqrt2}$
or, 1 - $\frac{\text{1 + }\sqrt2}{2\sqrt2}$ = sin2$\frac{45}{2}$
or, $\frac{2\sqrt2\text{- 1 - }\sqrt2}{2\sqrt2}$ = sin2$\frac{45}{2}$
or, $\frac{\sqrt2\text{ - 1}}{2\sqrt2}$ = sin2$\frac{45}{2}$
or, $\frac{\sqrt2\text{ - 1}}{2\sqrt2}$ × $\frac{\sqrt2}{\sqrt2}$ = sin2$\frac{45}{2}$
or, $\frac{2\text{ - }\sqrt2}{4}$ = sin2$\frac{45}{2}$
∴ sin$\frac{45}{2}^°$ = $\frac{\sqrt{2 - \sqrt{2}}}{2}$

Find the value of sin15° if cos30° = $\frac{\sqrt{3}}{2}$.

Here,
cos30° = $\frac{\sqrt{3}}{2}$

Now, cos30° = $\frac{\sqrt{3}}{2}$ or, cos2.15° = $\frac{\sqrt{3}}{2}$
or, 1 - 2sin215° = $\frac{\sqrt{3}}{2}$
or, 1 - $\frac{\sqrt{3}}{2}$ = 2sin215°
or, $\frac{2 \, - \, \sqrt{3}}{2}$ = 2sin215°
or, $\frac{2 \, - \, \sqrt{3}}{4}$ = sin215° ∴ sin15° = $\frac{\sqrt{\text{2 - }\sqrt{3}}}{2}$

If sin$\frac α3$ = $\frac 12$, find the value of sinα.

Here, sin$\frac α3$ = $\frac 12$

Now,
sinα = sin3.$\frac α3$ = 3sin$\frac α3$ - 4sin3$\frac α3$
= $\frac32$ - 4$\left(\frac 12\right)^3$
= $\frac 32$ - $\frac 48$
= 1

Prove that: $\frac{\text{1 + cosθ}}{sinθ}$ = cot$\frac θ2$

Here,
$\frac{\text{1 + cosθ}}{sinθ}$ = cot$\frac θ2$

Taking L.H.S,
= $\frac{\text{1 + cosθ}}{sinθ}$
= $\frac{\text{1 + }cos2.\frac θ2}{sin2.\frac θ2}$
= $\frac{2cos^2\fracθ2}{2sin\frac θ2 cos\frac θ2}$

∴ cot$\frac θ2$ = R.H.S