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Angles x and y are located in the first quadrant such that sinx=4/5, and cosy = 7/25.
a) Determine an exact value of cosx.
b) Determine an exact value of siny.
c) Determine an exact value for sin(x+y)
I'm really at a loss here, do I just sub those in to some formula or something? Can anyone help me out? Or get me pointed in the right direction? thanks in advance!!
This is a $\displaystyle 3-4-5$ triangle. If $\displaystyle \sin x = \frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}=\fra c{4}{5}$ then what is $\displaystyle \cos x = \frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}} = ...$.Quote:
Originally Posted by Random-Hero-
This is a $\displaystyle 7-24-25$ triangle. If $\displaystyle \cos y = \frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}} = \frac{7}{25}$ then what is $\displaystyle \sin y = \frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}= ...$.Quote:
Originally Posted by Random-Hero-
$\displaystyle \sin(x+y) = \sin x\cos y + \sin y \cos x$ thus once you have values of $\displaystyle \sin x, \cos y, \sin y, \cos x$, you can solve $\displaystyle \sin(x+y)$.Quote:
Originally Posted by Random-Hero-
Hello Random-HeroThe formulae you need here are:Quote:
Originally Posted by Random-Hero-
$\displaystyle \cos^2 A + \sin^2\ A = 1$ and $\displaystyle \sin(A + B) = \sin A\cos B + \cos A \sin B$
So in (a), use $\displaystyle \cos^2 x = 1 - \sin^2x$, and remember when you take the square root to find $\displaystyle \cos x$, that, since the angle $\displaystyle x$ is in the first quadrant $\displaystyle \cos x > 0$, so you don't have to worry about the negative square root.
(b) Use $\displaystyle \sin^2y = 1 - cos^2y$. And, again, since $\displaystyle y$ is in the first quadrant, $\displaystyle \sin y > 0$.
(c) simply use the formula for $\displaystyle \sin(A+B)$ and your answers to (a) and (b).
Can you complete it now?
Grandad
