Trignometric Ratios, Functions and Equations Test - 7

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

tan2x = tan(x+x)
(tanx + tanx)/(1 - tan² x)
= 2tanx/(1 - tan² x)

tan3 θ= (3tan θ- tan³ θ)/(1 - 3tan² θ)
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 2 θ= 2cos² θ- 1
= 2(1/√5)² - 1
= 2/5 - 1
= -3/5

cos θ= 1/2 (a + 1/a)
⇒cos 2 θ= 2 cos² θ−1
⇒cos 2 θ= 2 [1/2 (a + 1/a)]² −1
⇒cos 2 θ= 2 [1/4 (a² + 1/a² + 2)] −1
⇒cos 2 θ= 1/2 [a² + 1/a² + 2] −1
⇒cos 2 θ= a² /2 + 1/2a² + 1 −1
⇒cos 2 θ= 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

cos2 θ= cos2 θ−sin2 θ
And: cos2 θ+ sin2 θ=1
we find:
cos2 θ=(1 −sin2 θ)−sin2 θ=1 −2sin2 θ
θ= θ/4
cos θ/2 = 1 −2sin² θ/4