Question

Given cos theta=4sqrt2/7 find sin theta

Answer 1

$\\ \rm\Rrightarrow cos\theta=\dfrac{4\sqrt{2}}{7}$

Now

$\\ \rm\Rrightarrow sin^2\theta=1-cos^2\theta$

$\\ \rm\Rrightarrow sin^2\theta=1-\dfrac{16(2)}{49}$

$\\ \rm\Rrightarrow sin^2\theta=1-\dfrac{32}{49}$

$\\ \rm\Rrightarrow sin^2\theta=\dfrac{17}{49}$

$\\ \rm\Rrightarrow sin\theta=\dfrac{\sqrt{17}}{7}$

Answer 2

Answer:

• $sin \theta=\sqrt{17}/7$

Step-by-step explanation:

Given

• $cos \theta=4\sqrt{2}/7$

Find

• $sin \theta$

Solution

Use identity:

• $sin^2x+cos^2x=1$

Substitute and find:

• $cos^2 \theta=(4\sqrt{2}/7)^2$
• $cos^2 \theta=16*2/49=32/49$

• $sin \theta=\sqrt{1-cos^2 \theta} = \sqrt{1-32/49} =\sqrt{17/49} =\sqrt{17}/7$