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