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# Is there and easier way to do trig identities?

Any hints or tips?

### 4 Answers

- EdwardLv 71 decade agoFavorite Answer
Yes

Observe

sin(x) cos(x) tan(x) cot(x) sec(x) csec(x)

1/sin(x)=csec(x)

1/cos(x)=sec(x)

1/tan(x) =cot(x)

also

sin(x)/cos(x)=tan(x)

sin^2(x)+cos^2(x)=1

These are the basic level 1 (just kidding but if you got that you got 40% of the bulk)

Double angle

sin(2x)=2sin(x)cos(x)

cos(2x)=cos^2(x)-sin^2(x)

Knowing that puts you into 65-70th percentile.

Then you use these to derive additional 10-15%

Oh yes!

What about

sin(x+y)=? see rteference or math tables

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- ironduke8159Lv 71 decade ago
Start on the side that looks the most complicated.

Use well know identities that help you to isolate to just a single trig function.

Just keep trying and you'll get it finally.

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- 1 decade ago
hard way-practise

clever way- (a+b)^3

=(a+b)*(a+b)*(a+b)

=we can multiply each to get the formula

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