Test: Multiple Angle Formula - JEE MCQ

sin θ/3 
Multiply and divide it by '2'
sin2(θ/3*2)  = sin2(θ/6)
As we know that sin2θ = 2sinθcosθ
=> 2 sin(θ/6) cos(θ/6)

tan2x = tan(x+x)
(tanx + tanx)/(1 - tan²x)
= 2tanx/(1 - tan²x)

tan3θ = (3tanθ - tan³θ)/(1 - 3tan²θ)
tan θ = 1/2
= [3(1/2) - (1/2)³]/[1 - 3(1/2)²]
= [3/2 - 1/8]/[1 - 3/4]
= [11/8]/[1/4]
= 11/2

cos 2θ = 2cos²θ - 1
= 2(1/√5)² - 1
= 2/5 - 1
= -3/5

cos θ = 1/2 (a + 1/a)
⇒ cos 2θ = 2 cos²θ − 1
⇒ cos 2θ = 2 [1/2 (a + 1/a)]² − 1
⇒ cos 2θ = 2 [1/4 (a² + 1/a² + 2)] − 1
⇒ cos 2θ = 1/2 [a² + 1/a² + 2] − 1
⇒ cos 2θ = a²/2 + 1/2a² + 1 − 1
⇒ cos 2θ = 1/2 [a² + 1/a²]

Identity

 tan(45°) = tan(45°/2 + 45°/2)
= 2tan(45°/2)/(1 - tan²(45°/2))
(Using expansion for tan(2x))
This implies, 1 = 2tan(45°/2)/(1 - tan²(45°/2))
Rearranging terms, tan²(45°/2) + 2tan(45°/2) - 1 = 0
Solving the quadratic equation x² + 2x - 1 = 0 gives
x = (√2 - 1) or (-√2 - 1)
But tan(45°/2) lies in the first quadrant, therefore it should be positive.
tan(45°/2) = (√2 - 1)

Squaring on both sides,you get sin^2x + 2sinxcosx + cos^2x = 1

which gives 1 + 2sinxcosx = 1 {since sin^2x +cos^2x = 1 for all x}

which means,2sinxcosx =0

which means either sinx =0 or cosx=0

so,the general solution is that for sinx=0 union cosx=0

cos2θ = cos2θ − sin2θ
And: cos2θ + sin2θ=1
we find:
cos2θ=(1−sin2θ)−sin2θ=1−2sin2θ
θ = θ/4
cosθ/2 = 1−2sin²θ/4