# Further Trigonometry

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Hi, I have this problem with one exercise. Its in the "Mathemathics Standard Level for the IB Diploma" book by Smedley and Wiseman (oxford is the publisher) - maybe someone will find this helpful Anyway, these are exercises 9 through 15 on page 259.

The content is like this:

In question 9 - 15 prove each of the given identities.

9. 2cos2x - cos2x (triple equal sign o,O) 1

10. 2cos3x + sin2xsinx (triple equal sign o,O)2cosx

15. 1/(cosx+sinx) + 1/(cosx-sinx) (triple equal sign o,O) 2cox/cos2x

How do I solve this?

Thank you for the trouble,

Matt

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Do you mean 9(2cos2x-cos2x)≡1 or 9*2cos2x-cos2x≡1?

Edited by max0005

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Nevermind, my (embarrassing) mistake. Anyhow:

2cos2x-cos2x=1

2cos2x=1+cos2x

We know that: 1-2sin2x=cos2x

Hence:

2cos2x=1+1-2sin2x+1

2(cos2x+sin2x)=2

Fundamental identity:

sin2x+cos2x=1

Hope this helped, you should be able to solve the rest! ;D

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9. 2cos2x - cos2x (triple equal sign o,O) 1

10. 2cos3x + sin2xsinx (triple equal sign o,O)2cosx

15. 1/(cosx+sinx) + 1/(cosx-sinx) (triple equal sign o,O) 2cox/cos2x

Its proofs, not solving...

9.

LHS

= 2cos2x - cos2x

=2cos2x - (2cos2x - 1)

= 1

LHS=RHS

QED

10.

LHS

= 2cos3x + sin2xsinx

= 2cos3x + 2sinxcosxsinx

= 2cos3x + 2sin2xcosx

= 2cos3x + 2(1-cos2x)cosx

= 2cos3x + (2- 2cos2x)cosx

= 2cos3x + 2cosx- 2cos3x

= 2cosx

LHS=RHS

QED

I dont know what 11 is because it says cox..

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