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Sub-multiple Angles

If sinA3\frac A3 = 12(a+1a)\frac 12 (a + \frac 1a), prove that sinA = −12(a3+1a3)-\frac 12 (a^3 + \frac {1}{a^3})

Given: sinA3\frac A3 = 12(a+1a)\frac 12 (a + \frac 1a)
To Prove: sinA = −12(a3+1a3)-\frac 12 (a^3 + \frac {1}{a^3})

Now,
sinA = sin3A3\frac A3 = 3sinA3\frac A3 - 4sin3A3\frac A3
= 3 × 12(a+1a)\frac 12 (a + \frac 1a) - 4 ×{12(a+1a)}3\{\frac 12 (a + \frac 1a)\}^3
= 32(a+1a)\frac 32 (a + \frac 1a) - 48(a+1a)3\frac 48 (a + \frac 1a)^3
= 32(a+1a)\frac 32 (a + \frac 1a) - 12{a3+1a3+3.a.1a(a+1a)}\frac 12 \{a^3 + \frac{1}{a^3} + 3.a.\frac 1a (a + \frac 1a) \}
= 32(a+1a)\frac 32 (a + \frac 1a) - 12(a3+1a3)−32(a+1a)\frac 12 (a^3 + \frac{1}{a^3}) - \frac 32 (a + \frac 1a)

∴ sinA = −12(a3+1a3)-\frac 12 (a^3 + \frac {1}{a^3})

If cosB3\frac B3 = 12(m+1m)\frac 12 (m + \frac 1m), prove that cosB = 12(m3+1m3)\frac 12 (m^3 + \frac {1}{m^3})

Given: cosB3\frac B3 = 12(m+1m)\frac 12 (m + \frac 1m)
To prove: cosB = 12(m3+1m3)\frac 12 (m^3 + \frac {1}{m^3})

Now,
cosB = cos3B3\frac B3 = 4cos3B3\frac B3 - 3cosB3\frac B3
= 4 ×{12(m+1m)}3\{\frac 12 (m + \frac 1m)\}^3 - 3 × 12(m+1m)\frac 12 (m + \frac 1m)
= 48(m+1m)3\frac 48 (m + \frac 1m)^3 - 32(m+1m)\frac 32 (m + \frac 1m)
= 12{m3+1m3+3.a.1m(m+1m)}\frac 12 \{m^3 + \frac{1}{m^3} + 3.a.\frac 1m (m + \frac 1m) \} - 32(a+1m)\frac 32 (a + \frac 1m)
= 12(m3+1m3)+32(m+1m)\frac 12 (m^3 + \frac{1}{m^3}) + \frac 32 (m + \frac 1m) - 32(m+1m)\frac 32 (m + \frac 1m)

∴ cosB = 12(m3+1m3)\frac 12 (m^3 + \frac {1}{m^3})

If tanA2\frac A2 = 34\frac{3}{4}, find the value of sinA.

Given: tanA2\frac A2 = 34\frac{3}{4}

Now,
sinA = sin2.A2\frac A2 = 2tanA21 + tan2A2\frac{2tan\frac A2}{\text{1 + }tan^2\frac{A}{2}}
= 2.341 +916\frac{2.\frac 34}{\text{1 +}\frac{9}{16}}
= 64\frac 64 × 1625\frac{16}{25}
= 2425\frac{24}{25}

If sinA3\frac A3 = 45\frac{4}{5}, find the value of sinA.

Given: sinA3\frac A3 = 45\frac{4}{5}

Now,
sinA = sin3A3\frac A3 = 3sinA3\frac A3 - 4sin3A3\frac A3
= 3 × 45\frac 45 - 4 ×(45)3\left(\frac 45\right)^3
= 125\frac{12}{5} - 256125\frac{256}{125}
= 44125\frac{44}{125}

Prove that: 1 - cosθ + sinθ1 + cosθ + sinθ\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}} =tanθ2\frac{θ}{2}

Here,
1 - cosθ + sinθ1 + cosθ + sinθ\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}} =tanθ2\frac{θ}{2}

Taking L.H.S,
= 1 - cos2.θ2 + sinθ1 + cos2.θ2 + sinθ\frac{\text{1 - }cos2.\fracθ2\text{ + sinθ}}{\text{1 + }cos2.\fracθ2\text{ + sinθ}}
= 2sin2θ2 + sin2.θ22cos2θ2 + sin2.θ2\frac{2sin^2\frac θ2 \text{ + } sin2.\frac θ2}{2cos^2\frac θ2 \text{ + } sin2.\fracθ2}
= 2sin2θ2 + 2sinθ2cosθ22cos2θ2 + 2sinθ2cosθ2\frac{2sin^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}{2cos^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}
= 2sinθ2(sinθ2 + cosθ2)2cosθ2(cosθ2 + sinθ2)\frac{2sin\frac θ2(sin\frac θ2 \text{ + }cos\frac θ2)}{2cos\frac θ2(cos\frac θ2 \text{ + }sin\frac θ2)} = 1 - cosθ + sinθ1 + cosθ + sinθ\frac{\text{1 - cosθ + sinθ}}{\text{1 + cosθ + sinθ}}

∴ tanθ2\frac θ2 = R.H.S

Prove that: sinθ2 + sinθ1 + cosθ2 + cosθ\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}} = tanθ2\frac{θ}{2}

Here,
sinθ2 + sinθ1 + cosθ2 + cosθ\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}} = tanθ2\frac{θ}{2}

Taking L.H.S,
= sinθ2 + sinθ1 + cosθ2 + cosθ\frac{sin\frac{θ}{2}\text{ + sinθ}}{\text{1 + }cos\frac{θ}{2}\text{ + cosθ}} = sinθ2 + sin2.θ21 + cos2.θ2 + cosθ2\frac{sin\frac{θ}{2}\text{ + }sin2.\frac θ2}{\text{1 + }cos2.\fracθ2\text{ + }cos\frac{θ}{2}}
= sinθ2 + 2sin.θ2cosθ22cos2θ2 + cosθ2\frac{sin\frac{θ}{2}\text{ + }2sin.\frac θ2 cos\frac θ2}{2cos^2\fracθ2\text{ + }cos\frac{θ}{2}}
= sinθ2(1 + 2cosθ2)cosθ2(2cosθ2 + 1)\frac{sin\frac{θ}{2}\text{(1 + }2cos\frac θ2)}{cos\frac θ2(2cos\fracθ2\text{ + 1)}}

∴ tanθ2\frac θ2 = R.H.S

Prove that: cosecθ - cotθ = tanθ2\frac{θ}{2}

Here,
cosecθ - cotθ = tanθ2\frac{θ}{2}

Taking L.H.S,
= cosecθ - cotθ
= 1sinθ\frac{1}{sinθ} - cosθsinθ\frac{cosθ}{sinθ}
= 1 - cosθsinθ\frac{\text{1 - cosθ}}{sinθ}
= 1 - cos2.θ2sin2.θ2\frac{\text{1 - }cos2.\frac θ2}{sin2.\frac θ2}
= 2sin2θ22sinθ2cosθ2\frac{2sin^2\frac θ2}{2sin\fracθ2 cos\fracθ2}

∴ tanθ2\frac θ2 = R.H.S

Prove that: cotA2\frac{A}{2} - tanA2\frac{A}{2} = 2cotA

Here,
cotA2\frac{A}{2} - tanA2\frac{A}{2} = 2cotA

Taking L.H.S,
= cosA2sinA2\frac{cos\frac A2}{sin\frac A2} - sinA2cosA2\frac{sin\frac A2}{cos\frac A2}
= cos2A2 - sin2A2sinA2cosA2\frac{cos^2\frac A2 \text{ - }sin^2\frac A2}{sin\frac A2 cos\frac A2}
= cos2.A2sinA2cosA2\frac{cos2.\frac A2}{sin\frac A2 cos\frac A2} × 22\frac 22
= 2cosAsin2.A2\frac{2cosA}{sin2.\frac A2}
= 2cosAsinA\frac{2cosA}{sinA}

∴ 2cotA = R.H.S

Prove that: 1 + cosθ + sinθ1 - cosθ + sinθ\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}} = cotθ2\frac{θ}{2}

Here,
1 + cosθ + sinθ1 - cosθ + sinθ\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}} = cotθ2\frac{θ}{2}

Taking L.H.S,
= 1 + cosθ + sinθ1 - cosθ + sinθ\frac{\text{1 + cosθ + sinθ}}{\text{1 - cosθ + sinθ}}
= 1 + cos2.θ2 + sinθ1 - cos2.θ2 + sinθ\frac{\text{1 + }cos2.\fracθ2\text{ + sinθ}}{\text{1 - }cos2.\fracθ2\text{ + sinθ}}
= 2cos2θ2 + sin2.θ22sin2θ2 + sin2.θ2\frac{2cos^2\frac θ2 \text{ + } sin2.\frac θ2}{2sin^2\frac θ2 \text{ + } sin2.\fracθ2}
= 2cos2θ2 + 2sinθ2cosθ22sin2θ2 + 2sinθ2cosθ2\frac{2cos^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}{2sin^2\frac θ2 \text{ + } 2sin\frac θ2 cos\frac θ2}
= 2cosθ2(cosθ2 + sinθ2)2sinθ2(sinθ2 + cosθ2)\frac{2cos\frac θ2(cos\frac θ2 \text{ + }sin\frac θ2)}{2sin\frac θ2(sin\frac θ2 \text{ + }cos\frac θ2)}

∴ cotθ2\frac θ2 = R.H.S

If cosα2\frac{α}{2} = 12\frac{1}{2}, find the value of sinα.

Here,
cosα2\frac{α}{2} = 12\frac{1}{2}

Now,
sinα = sin2.α2\frac α2 = 2sinα2\frac α2.cosα2\frac α2
= 2 × 1 - cos2α2\sqrt{\text{1 - }cos^2\frac α2} × cosα2\frac α2
= 2 × 1 - 14\sqrt{\text{1 - }\frac 14} × 12\frac 12
= 4−14\sqrt{\frac{4 - 1}{4}}
= 32\frac{\sqrt{3}}{2}

If cosα3\frac{α}{3} = 12\frac{1}{2}, find the value of sinα.

Here,
cosα3\frac{α}{3} = 12\frac{1}{2}

Now,
sinα = sin3.α3\frac α3 = 3sinα3\frac α3 - 4sin3α3\frac α3
= 31 - cos2α3\sqrt{\text{1 - }cos^2\frac α3} - 4(1 - cos2α3)3\left(\sqrt{\text{1 - }cos^2\frac α3}\right)^3
= 31 - 14\sqrt{\text{1 - }\frac 14} - 4(1 - 14)3\left(\sqrt{\text{1 - }\frac 14}\right)^3
= 3 × 34\sqrt{\frac 34} - 4 × 343434\sqrt{\frac 34} \sqrt{\frac 34} \sqrt{\frac 34}
= 332\frac{3\sqrt{3}}{2} - 4 × 34\frac 34 × 32\frac{\sqrt3}{2} = 0

If cos45° = 12\frac{1}{\sqrt{2}}, prove that: sin(2212)\left(22\dfrac 12\right)° = 2−22\frac{\sqrt{2 - \sqrt{2}}}{2}

Given: cos45° = 12\frac{1}{\sqrt{2}}
To prove: sin452°\frac{45}{2}^° = 2−22\frac{\sqrt{2 - \sqrt{2}}}{2}

Now,
cos45° = 12\frac{1}{\sqrt{2}} or, cos2.452\frac{45}{2} = 12\frac{1}{\sqrt2}
or, 2cos2452\frac{45}{2} = 12\frac{1}{\sqrt2} + 1
or, 2(1 - sin2452)\left(\text{1 - }sin^2\frac{45}{2}\right) = 1 + 22\frac{\text{1 + }\sqrt2}{\sqrt2}
or, 1 - sin2452\text{1 - }sin^2\frac{45}{2} = 1 + 222\frac{\text{1 + }\sqrt2}{2\sqrt2}
or, 1 - 1 + 222\frac{\text{1 + }\sqrt2}{2\sqrt2} = sin2452\frac{45}{2}
or, 22- 1 - 222\frac{2\sqrt2\text{- 1 - }\sqrt2}{2\sqrt2} = sin2452\frac{45}{2}
or, 2 - 122\frac{\sqrt2\text{ - 1}}{2\sqrt2} = sin2452\frac{45}{2}
or, 2 - 122\frac{\sqrt2\text{ - 1}}{2\sqrt2} × 22\frac{\sqrt2}{\sqrt2} = sin2452\frac{45}{2}
or, 2 - 24\frac{2\text{ - }\sqrt2}{4} = sin2452\frac{45}{2}
∴ sin452°\frac{45}{2}^° = 2−22\frac{\sqrt{2 - \sqrt{2}}}{2}

Find the value of sin15° if cos30° = 32\frac{\sqrt{3}}{2}.

Here,
cos30° = 32\frac{\sqrt{3}}{2}

Now, cos30° = 32\frac{\sqrt{3}}{2} or, cos2.15° = 32\frac{\sqrt{3}}{2}
or, 1 - 2sin215° = 32\frac{\sqrt{3}}{2}
or, 1 - 32\frac{\sqrt{3}}{2} = 2sin215°
or, 2 − 32\frac{2 \, - \, \sqrt{3}}{2} = 2sin215°
or, 2 − 34\frac{2 \, - \, \sqrt{3}}{4} = sin215° ∴ sin15° = 2 - 32\frac{\sqrt{\text{2 - }\sqrt{3}}}{2}

If sinα3\frac α3 = 12\frac 12, find the value of sinα.

Here, sinα3\frac α3 = 12\frac 12

Now,
sinα = sin3.α3\frac α3 = 3sinα3\frac α3 - 4sin3α3\frac α3
= 32\frac32 - 4(12)3\left(\frac 12\right)^3
= 32\frac 32 - 48\frac 48
= 1

Prove that: 1 + cosθsinθ\frac{\text{1 + cosθ}}{sinθ} = cotθ2\frac θ2

Here,
1 + cosθsinθ\frac{\text{1 + cosθ}}{sinθ} = cotθ2\frac θ2

Taking L.H.S,
= 1 + cosθsinθ\frac{\text{1 + cosθ}}{sinθ}
= 1 + cos2.θ2sin2.θ2\frac{\text{1 + }cos2.\frac θ2}{sin2.\frac θ2}
= 2cos2θ22sinθ2cosθ2\frac{2cos^2\fracθ2}{2sin\frac θ2 cos\frac θ2}

∴ cotθ2\frac θ2 = R.H.S

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