Question

Find the exact value of the expression. sin(cos^−1 (3/4) − tan^−1(1/4))

Answer 1

do you know the trig SUM identities?

Answer 2

yes it sinu cosv + or - sinv cosu

Answer 3

yes.... well.. for this case is sin(a-b), so it'll end up as \(\bf sin(a)cos(b)-cos(a)sin(b)\)

so\(\bf sin\left[cos^{-1}\left(\frac{3}{4}\right)-tan^{-1}\left(\frac{1}{4}\right)\right]\implies sin(\theta-\beta)\\ \quad \\\implies sin(\theta)cos(\beta)cos(\theta)sin(\beta)\)

Answer 4

hmm... I missed the .... -... anyhow so \(\bf sin\left[cos^{-1}\left(\frac{3}{4}\right)-tan^{-1}\left(\frac{1}{4}\right)\right]\implies sin(\theta-\beta)\\ \quad \\
\implies sin(\theta)cos(\beta)-cos(\theta)sin(\beta)\)

so notice, each inverse function inside the SINE function, will return an angle, whatever angle that happens to be
in the end, the SUM or SUBTRACTION in this case, of both of those angles, will be a combination of the products of sine and cosine for each

Answer 5

so let's take a peek at both inverse functions
\(\bf sin\left[cos^{-1}\left(\frac{3}{4}\right)-tan^{-1}\left(\frac{1}{4}\right)\right]\\ \quad \\
-----------------------\\
cos^{-1}\left(\frac{3}{4}\right)\implies \theta\implies cos(\theta)=\cfrac{3}{4}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a}{c}\\ \quad \\
sin(\theta)=\cfrac{b}{c}\qquad c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\\
-----------------------\\
tan^{-1}\left(\frac{1}{4}\right)\implies \beta\implies tan(\beta)=\cfrac{1}{4}\implies \cfrac{opposite}{adjacent}\implies \cfrac{b}{a}\\ \quad \\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\\ \quad \\
cos(\beta)=\cfrac{a}{c}\qquad sin(\beta)=\cfrac{b}{c}\)

Answer 6

so all you need to do is really, find the missing side, from both angles, to find the sine or cosine as needed, and plug that in the SUM identity for sine

Answer 7

lemme find the 1st one, so you can find the next ones
\(\bf sin\left[cos^{-1}\left(\frac{3}{4}\right)-tan^{-1}\left(\frac{1}{4}\right)\right]\\ \quad \\
-----------------------\\
cos^{-1}\left(\frac{3}{4}\right)\implies \theta\implies cos(\theta)=\cfrac{3}{4}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a}{c}\\ \quad \\
sin(\theta)=\cfrac{b}{c}\qquad c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\implies \sqrt{4^2-3^2}=b=\sqrt{7}\\ \quad \\
sin(\theta)=\cfrac{\sqrt{7}}{4}\\ \quad \\ \quad \\
\textit{so it'd look like}\\
sin(\theta-\beta)=sin(\theta)cos(\beta)-cos(\theta)sin(\beta)\\ \quad \\
\implies sin(\theta-\beta)=\cfrac{\sqrt{7}}{4}\cdot cos(\beta)-\cfrac{3}{4}\cdot sin(\beta)\)