Question

Solve using identities​

Answer 1

• - 2√14/15

Explanation:

In the quadrant III both the sine and cosine get negative value.

Use the identity:

• sin²θ + cos²θ = 1

And consider negative value as mentioned above:

• sinθ = - √(1 - cos²θ)
• sinθ = - √(1 - (-13/15)²)
• sinθ = - √(1 - 169/225)
• sinθ = - √(56/225)
• sinθ = - 2√14/15

Answer 2

Solution given

Cos
`\displaystyle \theta_(1)=(13)/(15)`

consider Pythagorean theorem

`\bold{Sin²\theta+Cos²\theta=1}`

Subtracting
`Cos²\theta`both side

`\displaystyle Sin²\theta=1-Cos²\theta`

doing square root on both side we get

`Sin\theta=√(1-Cos²\theta)`

Similarly

`Sin\theta_(1)=\sqrt{1-Cos²\theta_(1)}`

Substituting value of
`Cos\theta_(1)`

we get

`Sin\theta_(1)=\sqrt{1-((-13)/(15))²}`

Solving numerical

`Sin\theta_(1)=\sqrt{1-((169)/(225))}`

`Sin\theta_(1)=\sqrt{(56)/(225)}`

`Sin\theta_(1)=(√(56))/(√(225))`

`Sin\theta_(1)=(√(2*2*14))/(√(15*15))`

`Sin\theta_(1)=(2√(14))/(15)`

Since

In III quadrant sin angle is negative

`\bold{Sin\theta_(1)=-(2√(14))/(15)}`