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Question
note: triangle may not be drawn to scale. suppose a = 10 and b = 8. find an exact value for each of the following trig functions. sin(a) = cos(a) = tan(a) = cot(a) = sec(a) = csc(a) =
Step1: Find the hypotenuse c
Use the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$. Given $a = 10$ and $b=8$, then $c=\sqrt{10^{2}+8^{2}}=\sqrt{100 + 64}=\sqrt{164}=2\sqrt{41}$.
Step2: Calculate $\sin(A)$
By the definition of sine in a right - triangle $\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}=\frac{10}{2\sqrt{41}}=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}$.
Step3: Calculate $\cos(A)$
By the definition of cosine in a right - triangle $\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c}=\frac{8}{2\sqrt{41}}=\frac{4}{\sqrt{41}}=\frac{4\sqrt{41}}{41}$.
Step4: Calculate $\tan(A)$
By the definition of tangent in a right - triangle $\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}=\frac{10}{8}=\frac{5}{4}$.
Step5: Calculate $\cot(A)$
Since $\cot(A)=\frac{1}{\tan(A)}$, then $\cot(A)=\frac{4}{5}$.
Step6: Calculate $\sec(A)$
Since $\sec(A)=\frac{1}{\cos(A)}$, then $\sec(A)=\frac{\sqrt{41}}{4}$.
Step7: Calculate $\csc(A)$
Since $\csc(A)=\frac{1}{\sin(A)}$, then $\csc(A)=\frac{\sqrt{41}}{5}$.
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$\sin(A)=\frac{5\sqrt{41}}{41}$
$\cos(A)=\frac{4\sqrt{41}}{41}$
$\tan(A)=\frac{5}{4}$
$\cot(A)=\frac{4}{5}$
$\sec(A)=\frac{\sqrt{41}}{4}$
$\csc(A)=\frac{\sqrt{41}}{5}$