Trig Equations: arcsin x = arccos(3/5) - arcsin(4/5), etc

[axrw]

I've got two problems that I'm not sure what to do on.

arcsin x = arccos ( 3/5) - arcsin ( 4/5)

and

Suppose that sin x = 5 cos x. Find sin x * cos x.

On the first I thought it would be something like:

x = cos( arccos(3/5) - arcsin( 4/5))

But I don't know what to do with that.

Thanks for any help.

EDIT: Ok, I just figured out the first one. I forgot about cosine sum identity. I still can't get the second though.

 

[Deleted member 4993]

Re: Trig Equations.

I've got two problems that I'm not sure what to do on.

arcsin x = arccos ( 3/5) - arcsin ( 4/5)

and

Suppose that sin x = 5 cos x. Find sin x * cos x.

tan x = 5

sec^2(x) = 26

cos^2(x) = 1/26

Now continue......

On the first I thought it would be something like:

x = cos( arccos(3/5) - arcsin( 4/5))

But I don't know what to do with that.

Thanks for any help.

EDIT: Ok, I just figured out the first one. I forgot about cosine sum identity. I still can't get the second though.

 

[axrw]

Thank you! Another identity I missed, I see. These identities will be the death of me.

 

[soroban]

Re: Trig Equations: arcsin x = arccos(3/5) - arcsin(4/5), et

Hello, axrw!

I caught an "eyeball" solution for the first one . . .

$\displaystyle \arcsin x \:= \:\underbrace{\arccos\left(\frac{3}{5}\right)}\, -\,\arcsin\left(\frac{4}{5}\right)$

Let $\displaystyle \arccos\left(\frac{3}{5}\right) \,=\,\theta\;\;\Rightarrow\;\;\cos\theta \,=\,\frac{3}{5} \,=\,\frac{adj}{hyp}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle \,adj \,=\,3,\;hyp\,=\,5$
From Pythagorus, we find that: $\displaystyle \,opp\,=\,4\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{4}{5}$
. . Hence: $\displaystyle \,\theta \,=\,\arcsin\left(\frac{4}{5}\right)$

The equation becomes: $\displaystyle \:\arcsin x \;=\;\arcsin\left(\frac{4}{5}\right)\,-\,\arcsin\left(\frac{4}{5}\right) \;=\;0$

Therefore: $\displaystyle \:x \:=\:0$

 

[axrw]

Thank you soroban. Your way is a bit more efficient than what I did.