Question

#QualityQuestion

@Trigonometry
$\bf{Prove ~the~ identity~:}Prove the identity : (1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)$
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Answer 1

Step-by-step explanation:

Straight away we notice that two trigonometric terms on the left hand side have 2x but there are no 2x on the right hand side therefore we realise that this question will require double angle formulae. If you can remember the double angle formulae that’s great, if not, you can easily obtain them from the trigonometric identities in your exam formula booklet: sin(A±B)=sinAcosB±cosAsinB cos(A±B)=cosAcosB∓sinAsinB (It’s important to note that the addition of A and B for cos correlates to the subtraction in the identity)

To work out the double angle formulae we simply make B=A and consider the addition case. This gives us the following: sin(2A)= sin(A+A)= sinAcosA+cosAsinA= 2sinAcosA cos(2A)= cos(A+A)= cosAcosA-sinAsinA= cos^2(A)-sin^2(A)

We also know that we can rewrite the right hand side in terms of sin(x) and cos(x) as follows: cot(x)=1/tan(x) = cos(x)/sin(x). This gives us an idea of what we are trying to make the left hand side look like.

These first two steps allow us to consider how we will go about proving the identity, they are not part of the proof themselves.

We are now ready to prove the identity. This is what your answer should look like:

Starting on the more complicated side, i.e. the left hand side, (1+cos(x)+cos(2x))/(sin(x)+sin(2x))=(1+cos(x)+cos^2(x)-sin^2(x))/(sin(x)+2sin(x)cos(x)) By substitution of the double angle formulae,

=(1+cos(x)+cos^2(x)-(1-cos^2(x)))/(sin(x)+2sin(x)cos(x)) By substitution of cos^2(x)+sin^2(x)=1 rearranged for sin^2(x)=1-cos^2(x),

=(cos(x)+2cos^2(x))/(sin(x)+2sin(x)cos(x)) By expanding and simplifying the numerator,

=(cos(x)(1+2cos(x)))/(sin(x)(1+2cos(x))) By factorising the numerator and the denominator,

=cos(x)/(sin(x)) By cancelling out (1+2cos(x)) as it is a like term,

=1/(

Answer 2

$\: \: \huge\fbox\pink{ᴀɴsᴡᴇʀ}$

➭ Given :-

(1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)

➭ Find :-

Prove the identity

➭ Soluction :-

To work out the double angle formulae we simply make B=A and consider the addition case.

This gives us the following:

sin(2A)= sin(A+A)= sinAcosA+cosAsinA= 2sinAcosA cos(2A)= cos(A+A)=

cosAcosA-sinAsinA= cos²(A)-sin²(A)

We also know that we can rewrite the right hand side in terms of sin(x) and cos(x) as follows: cot(x)=1/tan(x) = cos(x)/sin(x). This gives us an idea of what we are trying to make the left hand side look like.

Starting on the more complicated side,

i.e. the left hand side

(1+cos(x)+cos(2x))/(sin(x)+sin(2x))=(1+cos(x)+cos²(x)-sin²(x))/(sin(x)+2sin(x)cos(x))

By substitution of the double angle formulae,

=(1+cos(x)+cos²(x)-(1-cos²(x)))/(sin(x)+2sin(x)cos(x)) By substitution of cos²(x)+sin²(x)=1 rearranged for sin²(x)=1-cos²(x),

=(cos(x)+2cos²(x))/(sin(x)+2sin(x)cos(x))

By expanding and simplifying the numerator,

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

By factorising the numerator and the denominator,

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

By cancelling out (1+2cos(x)) as it is a like term,

=1/(tan(x))

By considering tan(x)=(sin(x))/(cos(x)),

=cot(x)

By definition,

Hence we have shown

(1+cos(x)+cos(2x))/(sin(x)+sin(2x))=cot(x) as required.

hope it was helpful to you