Trig Trick

Geometry Level 2

Multiple Choice Question

Simplify:

sin ⁡ 7 x − sin ⁡ 5 x cos ⁡ 7 x + cos ⁡ 5 x . \frac { \sin 7x - \sin 5x} { \cos 7x + \cos 5x}.

1) tan ⁡ x \tan x
2) tan ⁡ 2 x \tan 2x
3) tan ⁡ 6 x \tan 6x
4) tan ⁡ 12 x \tan 12x

[Sorry, they want the multiple choice options to be a whole number, so pick from the above choices.]

2 4 3 1

Chung Kevin by Chung Kevin , 45, Australia

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5 solutions

Before solving this problem, we must know some formulae: 1. sin ⁡ C − sin ⁡ D = 2 cos ⁡ C + D 2 × sin ⁡ C − D 2 1. \sin C-\sin D=2 \cos \frac{C+D}{2} \times \sin \frac{C-D}{2} 2. cos ⁡ C + cos ⁡ D = 2 cos ⁡ C + D 2 × cos ⁡ C − D 2 2. \cos C+\cos D=2 \cos \frac{C+D}{2} \times \cos \frac{C-D}{2} Now, let's solve the problem using these two formulae. sin ⁡ 7 x − sin ⁡ 5 x cos ⁡ 7 x + cos ⁡ 5 x \frac{\sin 7x-\sin 5x}{\cos 7x+\cos 5x} = cos ⁡ 7 x + 5 x 2 × sin ⁡ 7 x − 5 x 2 cos ⁡ 7 x + 5 x 2 × cos ⁡ 7 x − 5 x 2 =\frac{\cos \frac{7x+5x}{2} \times \sin \frac{7x-5x}{2}}{\cos \frac{7x+5x}{2} \times \cos \frac{7x-5x}{2}} = cos ⁡ 12 x 2 × sin ⁡ 2 x 2 cos ⁡ 12 x 2 × cos ⁡ 2 x 2 =\frac{\cos \frac{12x}{2} \times \sin \frac{2x}{2}}{\cos \frac{12x}{2} \times \cos \frac{2x}{2}} = cos ⁡ 6 x × sin ⁡ x c o s 6 x × cos ⁡ x =\frac{\cos 6x \times \sin x}{cos 6x \times \cos x} = sin ⁡ x cos ⁡ x =\frac{\sin x}{\cos x} = tan ⁡ x =\boxed{\tan x} As, among the four options, option 1 \boxed{1} has the correct answer i.e tan ⁡ x \boxed{\tan x} ,it is the correct option.

24 Helpful 0 Interesting 0 Brilliant 0 Confused

I think the right hand sides of the two formulae should be multiplied by 2.

Lauren Yeo - 7 years, 5 months ago

Hi can you please explain systematically how to use the formatting guide. am still having troubles writing out formulae like the you did. thanks

Abubakarr Yillah - 7 years, 5 months ago

Hello Abubakkar, the most important to remember while using the formatting guide is to enclose the Maths within braces and frontslashes. This is given in the formatting guide. Try this tip and Ithink your formulae will be okay. Please let me know if it worked or not.

Soham Dibyachintan - 7 years, 5 months ago

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well i have already tried putting them in brackets but nothing changes.. maybe i have been doing it wrongly..can you please demonstrate with any formula..so that i can see how it is done..thanks

Abubakarr Yillah - 7 years, 5 months ago

Yes, I had forggotten to do that as I was typing in a hurry, Thanks for reminding me.

Soham Dibyachintan - 7 years, 5 months ago
Anas M
Dec 15, 2013
  • sin7x - sin5x = 2 cos6x sinx
  • cos7x+cos5x = 2cos6x cosx
  • (sin7x - sin5x)/(cos7x+cos5x) = tanx
2 Helpful 0 Interesting 0 Brilliant 0 Confused

These formulae are required 1) sin(A+B) - sin(A-B) = 2cosAsinB 2) cos(A+B) + cos(A-B) = 2cosAcosB

The solution:--

[sin7x - sin5x] / [cos7x + cos5x] = [sin(6x+x) - sin(6x-x)] / [cos(6x+x) + cos(6x-x)] = [ 2 cos6x sinx ] / [ 2 cos6x cosx ] = sinx / cosx = tanx

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Abhishek Poman
Dec 16, 2013

(2cos(7x+5x/2) sin(7x-5x/2)) / (2cos(7x+5x/2) cos(7x-5x/2))

cancelling . .we get

sin(7x-5x/2) / cos(7x-5x/2)

sinx / cosx

=tanx

0 Helpful 0 Interesting 0 Brilliant 0 Confused

ok

Shahnawaz Kaka - 7 years, 5 months ago
Steve Mc
Dec 14, 2013

sin7x-sin5x=2sinxcos6x cos7x+cos5x=2cosxcos6x 2sinxcos6x/2cosxcos6x=tgx we canceled 2 and cos6x because of being commn

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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