Hinuku Posted May 24, 2011 Report Share Posted May 24, 2011 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 Reply Link to post Share on other sites More sharing options...
max0005 Posted May 24, 2011 Report Share Posted May 24, 2011 (edited) Do you mean 9(2cos2x-cos2x)≡1 or 9*2cos2x-cos2x≡1? Edited May 24, 2011 by max0005 Reply Link to post Share on other sites More sharing options...
max0005 Posted May 24, 2011 Report Share Posted May 24, 2011 Nevermind, my (embarrassing) mistake. Anyhow:2cos2x-cos2x=12cos2x=1+cos2xWe know that: 1-2sin2x=cos2xHence:2cos2x=1+1-2sin2x+12(cos2x+sin2x)=2Fundamental identity:sin2x+cos2x=1Hope this helped, you should be able to solve the rest! ;D 1 Reply Link to post Share on other sites More sharing options...
Peanut Butter Jelly Posted May 30, 2011 Report Share Posted May 30, 2011 9. 2cos2x - cos2x (triple equal sign o,O) 110. 2cos3x + sin2xsinx (triple equal sign o,O)2cosx15. 1/(cosx+sinx) + 1/(cosx-sinx) (triple equal sign o,O) 2cox/cos2xIts proofs, not solving...9. LHS= 2cos2x - cos2x=2cos2x - (2cos2x - 1)= 1LHS=RHS QED10.LHS= 2cos3x + sin2xsinx= 2cos3x + 2sinxcosxsinx= 2cos3x + 2sin2xcosx= 2cos3x + 2(1-cos2x)cosx= 2cos3x + (2- 2cos2x)cosx= 2cos3x + 2cosx- 2cos3x= 2cosxLHS=RHS QEDI dont know what 11 is because it says cox.. Reply Link to post Share on other sites More sharing options...
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