Question

Solve using identities​

Answer 1

Answer:

Solution given

Cos$\displaystyle \theta_{1}=\frac{13}{15}$

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

$Sin\theta=\sqrt{1-Cos²\theta}$

Similarly

$Sin\theta_{1}=\sqrt{1-Cos²\theta_{1}}$

Substituting value of $Cos\theta_{1}$

we get

$Sin\theta_{1}=\sqrt{1-(\frac{-13}{15})²}$

Solving numerical

$Sin\theta_{1}=\sqrt{1-(\frac{169}{225})}$

$Sin\theta_{1}=\sqrt{\frac{56}{225}}$

$Sin\theta_{1}=\frac{\sqrt{56}}{\sqrt{225}}$

$Sin\theta_{1}=\frac{\sqrt{2*2*14}}{\sqrt{15*15}}$

$Sin\theta_{1}=\frac{2\sqrt{14}}{15}$

Since

In III quadrant sin angle is negative

$\bold{Sin\theta_{1}=-\frac{2\sqrt{14}}{15}}$

Answer 2

Answer:

• - 2√14/15

Step-by-step 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