Auxillary method question (1 Viewer)

Lith_30 · Dec 3, 2021

How would I solve part a and b.

ExtremelyBoredUser · Dec 3, 2021

Lith_30 said:
View attachment 34400
How would I solve part a and b.

for a.
$\sin(x) + \cos(x) = \cos(2x)$
$\sin(x) + \cos(x) = cos^2x - sin^2x$
$\sin(x) + \cos(x) = (cosx+sinx)(cosx-sinx)$
$\sin(x) + \cos(x) - ( \sin(x) + \cos(x))( \cos(x) - \sin(x) ) = 0$
$(\sin(x) + \cos(x))(1 - \cos(x) + \sin(x))= 0$

and ye should be pre straight forward from there

b.
solve tanx = -1 and cosx - sinx = 1

ps. sorry for latex cbf

5uckerberg · Dec 3, 2021

For part a what can we turn $\cos{2\theta}$ into?
Next question we have $\cos^{2}\theta-\sin^{2}\theta=\sin{\theta}+\cos{\theta}$
we can turn that into $\cos^{2}\theta-\cos{\theta}=\sin{\theta}+\sin^{2}\theta$ . This becomes $\cos{\theta}(\cos{\theta}-1)=\sin{\theta}(1+\sin{\theta})$
part b is just using these two values and the second one is just the auxillary angle method. A useful technique there would be to draw out a right angled triangle with R being the hypotenuse and then finding the angle between the little sides and then you are done because the last step is simple simplifying.

Lith_30 · Dec 3, 2021

ExtremelyBoredUser said:
for a.
$\sin(x) + \cos(x) = \cos(2x)$
$\sin(x) + \cos(x) = cos^2x - sin^2x$
$\sin(x) + \cos(x) = (cosx+sinx)(cosx-sinx)$
$\sin(x) + \cos(x) - ( \sin(x) + \cos(x))( \cos(x) - \sin(x) ) = 0$
$(\sin(x) + \cos(x))(1 - \cos(x) + \sin(x))= 0$

and ye should be pre straight forward from there

b.
solve tanx = -1 and cosx - sinx = 1

ps. sorry for latex cbf

5uckerberg said:
For part a what can we turn $\cos{2\theta}$ into?
Next question we have $\cos^{2}\theta-\sin^{2}\theta=\sin{\theta}+\cos{\theta}$
we can turn that into $\cos^{2}\theta-\cos{\theta}=\sin{\theta}+\sin^{2}\theta$ . This becomes $\cos{\theta}(\cos{\theta}-1)=\sin{\theta}(1+\sin{\theta})$
part b is just using these two values and the second one is just the auxillary angle method. A useful technique there would be to draw out a right angled triangle with R being the hypotenuse and then finding the angle between the little sides and then you are done because the last step is simple simplifying.

Thanks guys, I get it now

Auxillary method question (1 Viewer)

Lith_30

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ExtremelyBoredUser

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5uckerberg

Well-Known Member

Lith_30

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