Question

Proving trigonometric identities. How to solve this using only the right side of the equation.

Answer 1

\[\sqrt{(5\cos(\theta)+12\sin(\theta))^2+(12\cos(\theta)-5\sin(\theta))^2}=13\]

Answer 2

I have already solved this by manipulating the left side but our teacher said that we could also solve it by manipulating only the right side.

Answer 3

work backwards from your answer on the left side and that's your answer for the right side

Answer 4

i mean, how do you make "13" equal to the left side of the equation?

Answer 5

how did you make the left side of the equation equal to 13?

Answer 6

@Auxuris first I simplified the terms \[\sqrt{(25\cos^2(\theta)+120\cos(\theta)\sin(\theta)+144\sin^2(\theta))+(144\cos^2(\theta)-120\cos(\theta)\sin(\theta)+25\sin^2(\theta)}\]

Answer 7

which became \[\sqrt{169\cos^2(\theta)+169\sin^2(\theta)}=13\]

then factoring out 169,
\[\sqrt{169(\cos^2(\theta)+\sin^2(\theta))}=13\]

since \[\cos^2(\theta)+\sin^2(\theta)=1\]

the equation will be \[\sqrt{169}=13\]

and the square root of 169 is 13, so 13=13

Answer 8

That is correct.

Answer 9

@blackbird02 please, try this:
\[13=13*1=13*[(\cos \theta)^{2}+(\sin \theta)^{2}]=13 (\cos \theta)^{2}+13(\sin \theta)^{2}\]

Answer 10

@Michele_Laino What do I do next?

Answer 11

Sorry, better may be this:
\[13=(\sqrt{13*1})^{2}=(\sqrt{13[(\cos \theta)^{2}+(\sin \theta)^{2}]})^{2}\]

Answer 12

next substitute, like above, and square both sides of your equality

Answer 13

sorry again,
\[13=\sqrt{169*1}=\sqrt{169*[(\cos \theta)^{2}+(\sin \theta)^{2}]}=...\]

Answer 14

now substitute the above quantity at the right side, and then square both sides of the resultant equality

Answer 15

@blackbird02 try please

Answer 16

Do I have to square the left side of the equation too?

Answer 17

yes! both sides, please!

Answer 18

Just a sec

Answer 19

of course, first substitute the right side, how I have indicated above please!

Answer 20

Will there be any other way with out manipulating the left side?

Answer 21

I try to find it

Answer 22

I'm just curious to know how to make the right side equal to the left side without touching or manipulating the left part of the equation. Our calculus teacher said there's a way but I don't know how.

Answer 23

maybe I found it:
please try to substitute to right side of your equality this:
\[13=\sqrt{169*[(\cos (\theta+\alpha))^{2}+(\sin(\theta+ \alpha))^{2}}\]
where alpha is such that:
\[\sin \alpha =\frac{ 5 }{ 13 }, \cos \alpha=\frac{ 12 }{ 13 }\]
then square both sides of your equality without manipulating the left side

Answer 24

@blackbird02 try please, I think it is the method of your teacher