Finding exact trig values from sine and cosine given with angles

Expert intro: This exercise combines reference triangles, Pythagorean identities, reciprocal identities, and double-angle and half-angle formulas.

Detailed walkthrough

Start with the given information

We are told that

$\sin(x)=\frac{8}{17}, \qquad \cos(y)=\frac{7}{25}.$

From these, we can build the rest of the trig values using the identity

$\sin^2\theta+\cos^2\theta=1.$

Because the question does not specify quadrants, the usual assumption in such exact-value problems is that the angles are in a quadrant where the indicated ratios determine the sign, or that the principal-triangle values are intended.

Find the ratios for $x$

Since $\sin(x)=\frac{8}{17}$, use a right triangle with opposite side 8 and hypotenuse 17. Then the adjacent side is

$\sqrt{17^2-8^2}=\sqrt{289-64}=15.$

So

$\cos(x)=\frac{15}{17},\quad \tan(x)=\frac{8}{15},\quad \csc(x)=\frac{17}{8},\quad \sec(x)=\frac{17}{15},\quad \cot(x)=\frac{15}{8}.$

Now use double-angle formulas:

$\sin(2x)=2\sin x\cos x=2\cdot\frac{8}{17}\cdot\frac{15}{17}=\frac{240}{289},$

$\cos(2x)=\cos^2x-\sin^2x=\left(\frac{15}{17}\right)^2-\left(\frac{8}{17}\right)^2=\frac{225-64}{289}=\frac{161}{289}.$

Find the ratios for $y$

Since $\cos(y)=\frac{7}{25}$, use a right triangle with adjacent side 7 and hypotenuse 25. Then the opposite side is

$\sqrt{25^2-7^2}=\sqrt{625-49}=24.$

Thus

$\sin(y)=\frac{24}{25},\quad \tan(y)=\frac{24}{7},\quad \csc(y)=\frac{25}{24},\quad \sec(y)=\frac{25}{7},\quad \cot(y)=\frac{7}{24}.$

For the half-angle values,

$\sin\left(\frac y2\right)=\sqrt{\frac{1-\cos y}{2}}=\sqrt{\frac{1-\frac{7}{25}}{2}}=\sqrt{\frac{9}{25}}=\frac35,$

and

$\cos\left(\frac y2\right)=\sqrt{\frac{1+\cos y}{2}}=\sqrt{\frac{1+\frac{7}{25}}{2}}=\sqrt{\frac{16}{25}}=\frac45.$

Final answers

(a)\ &\cos x=\frac{15}{17} &\qquad (j)\ &\sin y=\frac{24}{25}\\ (b)\ &\tan x=\frac{8}{15} &\qquad (k)\ &\tan y=\frac{24}{7}\\ (c)\ &\csc x=\frac{17}{8} &\qquad (l)\ &\csc y=\frac{25}{24}\\ (d)\ &\sec x=\frac{17}{15} &\qquad (m)\ &\sec y=\frac{25}{7}\\ (e)\ &\cot x=\frac{15}{8} &\qquad (n)\ &\cot y=\frac{7}{24}\\ (f)\ &\sin(2x)=\frac{240}{289} &\qquad (o)\ &\sin\left(\frac y2\right)=\frac35\\ (g)\ &\cos(2x)=\frac{161}{289} &\qquad (p)\ &\cos\left(\frac y2\right)=\frac45 \end{aligned}$$ The fastest route is to convert each given ratio into a right triangle, then use identities rather than trying to memorize every value separately. ### 💡 Pitfall guide A frequent error is using the half-angle formulas without checking the sign. The expressions $\sqrt{\frac{1\pm\cos y}{2}}$ give the magnitude, but the sign depends on the quadrant of $y/2$. In many textbook settings, the intended answer is the positive principal value, but on a real exam you should verify the quadrant if it is given. Another common mistake is mixing up $\tan\theta=\frac{\sin\theta}{\cos\theta}$ with $\frac{\cos\theta}{\sin\theta}$. If you build the triangle carefully, these reciprocal ratios stay consistent. ### 🔄 Real-world variant If instead the problem gave $\sin(x)=\frac{8}{17}$ and $\cos(y)=-\frac{7}{25}$, then the triangle for $y$ would lie in a quadrant where cosine is negative, and the signs of $\sin y$, $\tan y$, and the half-angle values would need quadrant analysis. For example, if $y$ were in Quadrant II, then $\sin y$ would still be positive, but $\tan y$ would be negative. That is why exact-value questions often pair a trig ratio with a quadrant clue. ### 🔍 Related terms Pythagorean identity, double-angle formula, half-angle formula