# root (a-asin(x) = cos(x) ? or what? [closed]

Can you explain whats going on here? is it some trig identity i dont know or something else?

## closed as off-topic by Shaun, Mohammad Riazi-Kermani, Xander Henderson, JMP, choco_addictedMar 16 '18 at 6:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Mohammad Riazi-Kermani, Xander Henderson, JMP, choco_addicted
If this question can be reworded to fit the rules in the help center, please edit the question.

• You're getting a lot of downvotes because you haven't supplied us with an attempt on your part. – Shaun Mar 16 '18 at 1:17
• – Shaun Mar 16 '18 at 1:17

Factor $R^2$ and note that

$$1- \sin ^2 (x) = \cos ^2 (x).$$

• but if i put (5-5sin(90))^0.5 in a calculator is not the same as 5 (1-1sin(90))^0.5 – tgmjack Mar 16 '18 at 1:11
• is the 5-5sin(90) not the two terms? – tgmjack Mar 16 '18 at 1:13
• @tgmjack $\sqrt{5-5\sin^2(90^\circ)} = \sqrt{5(1-\sin^2(90^\circ))}$ – Andrew Li Mar 16 '18 at 1:15

Try this:

$\sqrt{R^2-R^2(\sin\theta)^2} = \sqrt{R^2(1-\sin^2 \theta)}$

And because

$1-\sin^2 (\theta) = \cos^2 (\theta)$

You can get

$\sqrt{R^2\cos^2 (\theta)} = R\cos(\theta)$

• yeah but it equals 2R^2*cos^2(x) not 2√R^2*cos^2(x) where does the root go? if the power cancels the roots – tgmjack Mar 16 '18 at 1:31
• the root cancels the powers – Dashi Mar 16 '18 at 3:14