Question

Please help me thank you

Answer 1

`\large\boxed{\sin2\theta=(\sqrt3)/(2),\ \cos2\theta=(1)/(2)}`

Explanation:

We know:

`\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet`

We have

`\sin\theta=(1)/(2)`

Use
`\sin^2\theta+\cos^2\theta=1`

`\left((1)/(2)\right)^2+\cos^2\theta=1\\\\(1)/(4)+\cos^2\theta=1\qquad\text{subtract}\ (1)/(4)\ \text{from both sides}\\\\\cos^2\theta=(4)/(4)-(1)/(4)\\\\\cos^2\theta=(3)/(4)\to\cos\theta=\pm\sqrt{(3)/(4)}\to\cos\theta=\pm(\sqrt3)/(\sqrt4)\to\cos\theta=\pm(\sqrt3)/(2)\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=(\sqrt3)/(2)`

`\sin2\theta=2\left((1)/(2)\right)\left((\sqrt3)/(2)\right)=(\sqrt3)/(2)\\\\\cos2\theta=\left((\sqrt3)/(2)\right)^2-\left((1)/(2)\right)^2=(3)/(4)-(1)/(4)=(3-1)/(4)=(2)/(4)=(1)/(2)`