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Introduction to Trigonometry

All 6 textbook questions with direct answer previews. Open any question for simple explanations and exam-ready answers.

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All questions

6
Q1

Define the six trigonometric ratios for an acute angle AA in a right triangle.

For angle AA: sin⁡A=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}}, cos⁡A=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, tan⁡A=oppositeadjacent\tan A=\frac{\text{opposite}}{\text{adjacent}}, cosec⁡A=1sin⁡A\cosec A=\frac1{\sin A}, sec⁡A=1cos⁡A\sec A=\frac1{\cos A}, cot⁡A=1tan⁡A\cot A=\frac1{\tan A}.
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Q2

Show that tan⁡A=sin⁡Acos⁡A\tan A=\frac{\sin A}{\cos A}.

Since sin⁡A=oppositehypotenuse\sin A=\frac{\text{opposite}}{\text{hypotenuse}} and cos⁡A=adjacenthypotenuse\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}, sin⁡Acos⁡A=oppositeadjacent=tan⁡A\frac{\sin A}{\cos A}=\frac{\text{opposite}}{\text{adjacent}}=\tan A.
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Q3

Write the values of sin⁡30∘\sin 30^\circ, cos⁡60∘\cos 60^\circ, tan⁡45∘\tan 45^\circ, and sec⁡60∘\sec 60^\circ.

sin⁡30∘=12\sin30^\circ=\frac12, cos⁡60∘=12\cos60^\circ=\frac12, tan⁡45∘=1\tan45^\circ=1, and sec⁡60∘=2\sec60^\circ=2.
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Q4

State the trigonometric ratios of complementary angles.

For acute AA, sin⁡(90∘−A)=cos⁡A\sin(90^\circ-A)=\cos A, cos⁡(90∘−A)=sin⁡A\cos(90^\circ-A)=\sin A, tan⁡(90∘−A)=cot⁡A\tan(90^\circ-A)=\cot A, cot⁡(90∘−A)=tan⁡A\cot(90^\circ-A)=\tan A, sec⁡(90∘−A)=cosec⁡A\sec(90^\circ-A)=\cosec A, and cosec⁡(90∘−A)=sec⁡A\cosec(90^\circ-A)=\sec A.
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Q5

Prove the identity sin⁡2A+cos⁡2A=1\sin^2 A+\cos^2 A=1.

In a right triangle, sin⁡A=ph\sin A=\frac{p}{h} and cos⁡A=bh\cos A=\frac{b}{h}. Then sin⁡2A+cos⁡2A=p2+b2h2=h2h2=1\sin^2A+\cos^2A=\frac{p^2+b^2}{h^2}=\frac{h^2}{h^2}=1 by Pythagoras theorem.
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Q6

Evaluate sin⁡230∘+cos⁡230∘\sin^2 30^\circ+\cos^2 30^\circ.

By the identity sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1, the value is 11. Directly, (12)2+(32)2=14+34=1(\frac12)^2+(\frac{\sqrt3}{2})^2=\frac14+\frac34=1.
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