Trig identity.Help

Need help in these two questions
Thanks in advance

First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA

Second question square both sides to start and then rewrite the cos2x term and start working towards something that could resemble tanx

Original post by mathcoachni

First part of first question from double angle identities, then it's a case of rearranging and factorising until your left with sinA/CosA = tanA

I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that

I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after

Remember what your trying to make the LHS look like, you want sin/cos, sometimes it just takes you to go through the different options.

I did this part so far already and with one more step which is factorising Cos^2(A)-Sin^2(A) but got stuck after that

remember that

cos(2A) \equiv cos^2A - sin^2A \equiv 2cos^2A - 1

this is done by using

sin^2A + cos^2A \equiv 1 \implies sin^2A \equiv 1 - cos^2A

now in the denominator, the -1 and +1 cancel out each other, so you are left with:

\frac{2sinAcosA + sinA}{2cos^2A + cosA}

Factorise this now and you should be able to get the answer

(edited 7 years ago)

Thanks alot mate

How to revise for GCSE Maths: AQA explains what to do

What is an LLM and how could it benefit your career?

Powerful questions business students forget to ask their career advisers

Why I chose a law degree and what studying law was like