Question

Given: cos(theta) = 0,6 ; 270° < theta < 360°

Find: 1) sin2(theta)
2) cos2(theta)
3) tan4(theta)
Additional tools usage solving this question is prohibited

Answer 1

Explanation:

`\sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha )`

To find sin(a),

`\sin( \alpha ) = \sqrt{1 - \cos {}^(2) ( \alpha ) }`

`\sin( \alpha ) = \sqrt{1 - (0.6) {}^(2) }`

`\sin( \alpha ) = √(0.64)`

Since sin is negative in the interval (270,360)

`\sin( \alpha ) = - 0.8`

Now we can find sin(2a)

`2( - 0.8)(0.6) = - 0.96`

2. The identity for cos(2x) is

`2 \cos {}^(2) (x) - 1`

We know cos x is 0.6 so we get

`2(0.6) {}^(2) - 1 = - 0.28`

part 3:

`\tan(4x) = \frac{2 \tan(2x) }{1 - \tan {}^(2) (2x) }`

`\tan(2x) = ( \sin( 2x) ) )/( \cos(2x) ) = ( - 0.96)/( - 0.28) = 24`

So we end up getting,

`\tan(4x) = \frac{2(24)}{1 - (24) {}^(2) }`

`\tan(4x) = (48)/(1 - 576)`

`\tan(4x) = - (48)/(575)`

Repost the question, I made a computational mistake.