Finding Related Trig Ratios | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Recover the missing side

Given $\cos\theta=-3/5$ in quadrant III, find sine and tangent.

Choose radius 5 and horizontal coordinate −3. Pythagoras gives the missing vertical coordinate:

y^2+(-3)^2=5^2

y^2=25-9=16

Both 4 and −4 square to 16. In III the point is below the axis, so choose $y=-4$ .

Use positive lengths in Pythagoras; choose signs from the quadrant.

Answer

\sin\theta=-\frac45

\tan\theta=\frac{-4}{-3}=\frac43

The triangle uses lengths 3, 4, 5; the coordinates supply the signs. Do not make a side length negative. If the question gives no quadrant, check whether both signs are possible.

The same method works with a letter

Suppose $\sin\alpha=k$ , with $\alpha$ acute. Choose hypotenuse 1 and opposite side $k$ . The adjacent side is $\sqrt{1-k^2}$ , the positive square root because a side length is positive.

\cos\alpha=\sqrt{1-k^2}

\tan\alpha=\frac{k}{\sqrt{1-k^2}}

Here $0<k<1$ . If the angle were in quadrant II instead, cosine and tangent would be negative.

The two acute angles of a right triangle add to 90° and are called . Switching between them swaps opposite and adjacent:

\sin(90^\circ-\alpha)=\cos\alpha

\cos(90^\circ-\alpha)=\sin\alpha

So if $\sin50^\circ=k$ , then $\cos40^\circ=k$ and $\tan40^\circ=\sqrt{1-k^2}/k$ . In radians, use $\pi/2$ instead of 90°.

Your turn

Your turnPractice

$\sin\beta=a$ and $90^\circ<\beta<180^\circ$ . Express $\cos\beta$ , $\tan\beta$ and $\sin(180^\circ-\beta)$ in terms of $a$ .

Show worked answer

Quadrant II has negative cosine and tangent. The reference angle $180^\circ-\beta$ has the same positive sine.

Answer

\cos\beta=-\sqrt{1-a^2}

\tan\beta=-\frac{a}{\sqrt{1-a^2}}

\sin(180^\circ-\beta)=a

The restriction implies $0<a<1$ , so the square root is real and the denominator is non-zero.

Recover the missing side

Given $\cos\theta=-3/5$ in quadrant III, find sine and tangent.

Choose radius 5 and horizontal coordinate −3. Pythagoras gives the missing vertical coordinate:

y^2+(-3)^2=5^2

y^2=25-9=16

Both 4 and −4 square to 16. In III the point is below the axis, so choose $y=-4$ .

Use positive lengths in Pythagoras; choose signs from the quadrant.

Answer

\sin\theta=-\frac45

\tan\theta=\frac{-4}{-3}=\frac43

The triangle uses lengths 3, 4, 5; the coordinates supply the signs. Do not make a side length negative. If the question gives no quadrant, check whether both signs are possible.

The same method works with a letter

Suppose $\sin\alpha=k$ , with $\alpha$ acute. Choose hypotenuse 1 and opposite side $k$ . The adjacent side is $\sqrt{1-k^2}$ , the positive square root because a side length is positive.

\cos\alpha=\sqrt{1-k^2}

\tan\alpha=\frac{k}{\sqrt{1-k^2}}

Here $0<k<1$ . If the angle were in quadrant II instead, cosine and tangent would be negative.

The two acute angles of a right triangle add to 90° and are called . Switching between them swaps opposite and adjacent:

\sin(90^\circ-\alpha)=\cos\alpha

\cos(90^\circ-\alpha)=\sin\alpha

So if $\sin50^\circ=k$ , then $\cos40^\circ=k$ and $\tan40^\circ=\sqrt{1-k^2}/k$ . In radians, use $\pi/2$ instead of 90°.

Your turn

Your turnPractice

$\sin\beta=a$ and $90^\circ<\beta<180^\circ$ . Express $\cos\beta$ , $\tan\beta$ and $\sin(180^\circ-\beta)$ in terms of $a$ .

Show worked answer

Quadrant II has negative cosine and tangent. The reference angle $180^\circ-\beta$ has the same positive sine.

Answer

\cos\beta=-\sqrt{1-a^2}

\tan\beta=-\frac{a}{\sqrt{1-a^2}}

\sin(180^\circ-\beta)=a

The restriction implies $0<a<1$ , so the square root is real and the denominator is non-zero.