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8 months ago

Prove that co3x / sinx + sin3x / cosx = 2cot2x

I looked at the mark scheme but I don't understand the 1st two steps:

(cos3xcosx + sin3x) / sinxcosx =

cos(3x-x)/sinxcosx =

...

Can someone please explain to me what they're doing in the 1st 2 steps? I thought the 1st step was multiplying by sinx and cosx so that the denominators are both sinxcosx but the markscheme doesn't show it being multiplied by sinx only cosx.

Thankyou

I looked at the mark scheme but I don't understand the 1st two steps:

(cos3xcosx + sin3x) / sinxcosx =

cos(3x-x)/sinxcosx =

...

Can someone please explain to me what they're doing in the 1st 2 steps? I thought the 1st step was multiplying by sinx and cosx so that the denominators are both sinxcosx but the markscheme doesn't show it being multiplied by sinx only cosx.

Thankyou

Original post by random_account

Prove that co3x / sinx + sin3x / cosx = 2cot2x

I looked at the mark scheme but I don't understand the 1st two steps:

(cos3xcosx + sin3x) / sinxcosx =

cos(3x-x)/sinxcosx =

...

Can someone please explain to me what they're doing in the 1st 2 steps? I thought the 1st step was multiplying by sinx and cosx so that the denominators are both sinxcosx but the markscheme doesn't show it being multiplied by sinx only cosx.

Thankyou

I looked at the mark scheme but I don't understand the 1st two steps:

(cos3xcosx + sin3x) / sinxcosx =

cos(3x-x)/sinxcosx =

...

Can someone please explain to me what they're doing in the 1st 2 steps? I thought the 1st step was multiplying by sinx and cosx so that the denominators are both sinxcosx but the markscheme doesn't show it being multiplied by sinx only cosx.

Thankyou

Must be a typo and they missed off the sin(x) on line 1 as the additive "identity" on line 2 assumes its there.

(edited 8 months ago)

Reply 2

8 months ago

Original post by mqb2766

Must be a typo and they missed off the sin(x) on line 1 as the additive "identity" on line 2 assumes its there.

Yh it might be but even then I don't understand the 2nd step. Would you mind explaining it to me? Thankyou

Original post by random_account

Yh it might be but even then I don't understand the 2nd step. Would you mind explaining it to me? Thankyou

Its the additive identity which you should have come across so

cos(A-B) = cos(A)cos(B)+sin(A)sin(B)

Tbh, you need to have a reasonable idea of what youre trying to do, so working back from the right hand side youre trying to get roughly

cos(2x) / sin(2x)

and this is done by having sin(x)cos(x) on the denominator which is ~sin(2x) and on the numerator you want to combine trig(3x) and trig(x) terms to get a cos(2x), and youd need to do an identity for cos(3x-x) to get this.

cos(3x) -> cos(2x + x)

sin(3x) -> sin(2x+x)

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

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

times sin(2x+x) by sin x, times cos(2x+x) by cos x, so that we can combine it into a single fraction.

now we have (cos(2x)cos^2x - sin(2x)sin(x)cos(x) + sin(2x)sin(x)cos(x) + cos(2x)sin^2x) / cos(x)sin(x)

the sin(2x)sin(x)cos(x) cancels out as we have a - then a +

this becomes just (cos(2x)cos^2x + cos(2x)sin^2x) / cos(x)sin(x)

factor out cos(2x) , becomes cos(2x)(cos^2x+sin^2x) / cos(x)sin(x)

cos^2x + sin^2x = 1

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

this becomes 2cos(2x)/sin(2x) == 2cot2x

sin(3x) -> sin(2x+x)

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

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

times sin(2x+x) by sin x, times cos(2x+x) by cos x, so that we can combine it into a single fraction.

now we have (cos(2x)cos^2x - sin(2x)sin(x)cos(x) + sin(2x)sin(x)cos(x) + cos(2x)sin^2x) / cos(x)sin(x)

the sin(2x)sin(x)cos(x) cancels out as we have a - then a +

this becomes just (cos(2x)cos^2x + cos(2x)sin^2x) / cos(x)sin(x)

factor out cos(2x) , becomes cos(2x)(cos^2x+sin^2x) / cos(x)sin(x)

cos^2x + sin^2x = 1

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

this becomes 2cos(2x)/sin(2x) == 2cot2x

Original post by Adamlewry

...

Best not to post full solutions, and the approach started in the OP is simpler/more direct

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