10E, 10F - Using trigonometric identities

10E, 10F - Using trigonometric identities

Knowing one trigonometric ratio often lets you find the others without finding the angle. In 10E, you will connect sine and cosine through the unit circle; in 10F, you will write tangent as their quotient. Together, these identities let you calculate exact values and prove that different expressions are equal wherever they are defined.

The Pythagorean identity

If you know that sin⁡θ=45\sin\theta=\frac{4}{5}, must cos⁡θ\cos\theta be 35\frac{3}{5}? The size is determined, but its sign depends on where the angle ends.

For example, the point (−35,45)\left(-\frac35,\frac45\right) lies on the unit circle because 925+1625=1\frac9{25}+\frac{16}{25}=1. Its horizontal coordinate is negative and its vertical coordinate is positive, so it represents an angle in the second quadrant. Its cosine is −35-\frac35, even though its sine is positive.

[DIAGRAM: asset_name: Sine and cosine on the unit circle; asset_slug: identity-unit-circle; caption: The radius is 1, so the squares of the horizontal and vertical coordinates add to 1; description: A unit circle centred at O has a radius to a point P in quadrant two, labelled P equals cos theta comma sin theta. A perpendicular from P to the horizontal axis makes a right triangle. The horizontal coordinate is negative and the vertical coordinate positive. A counterclockwise arc from the positive horizontal axis marks theta. Alongside, Pythagoras gives cos squared theta plus sin squared theta equals 1.]
The radius is 1, so the squares of the horizontal and vertical coordinates add to 1

On a unit circle, the point at angle θ\theta has coordinates (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Pythagoras gives the same equation in every quadrant: squaring a negative coordinate makes it positive. It also holds on the axes, where one coordinate is zero.

Pythagorean identity

For every real angle θ\theta, in degrees or radians,

sin⁡2θ+cos⁡2θ=1.\sin^2\theta+\cos^2\theta=1.

Here sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2, not sin⁡(θ2)\sin(\theta^2).

Not on the formulae sheet: learn it.

An identity is an equality that holds for every value in its stated domain. Use this one to find a missing squared ratio, then use the quadrant to choose its sign.

Worked example 1: an acute angle

Question. Given sin⁡θ=513\sin\theta=\frac5{13} and 0∘<θ<90∘0^\circ<\theta<90^\circ, find the exact value of cos⁡θ\cos\theta.

Plan. Find cos⁡2θ\cos^2\theta, then choose the positive root because the angle is acute.

cos⁡2θ=1−sin⁡2θrearrange the identity=1−25169=144169substitute the given sine\begin{aligned} \cos^2\theta&=1-\sin^2\theta &&\text{rearrange the identity}\\ &=1-\frac{25}{169}=\frac{144}{169} &&\text{substitute the given sine} \end{aligned}

Both roots have size 1213\frac{12}{13}. In the first quadrant cosine is positive.

Answer. cos⁡θ=1213\cos\theta=\frac{12}{13}.

Check. A right triangle with opposite side 55 and hypotenuse 1313 has adjacent side 132−52=12\sqrt{13^2-5^2}=12, giving the same cosine.

Keep the same sine but move the angle to the second quadrant. Only the sign of cosine changes.

Worked example 2: the other possible sign

Question. Given sin⁡θ=513\sin\theta=\frac5{13} and 90∘<θ<180∘90^\circ<\theta<180^\circ, find cos⁡θ\cos\theta exactly.

Plan. Use the identity for the magnitude and the second quadrant for the sign.

cos⁡2θ=1−(513)2=144169.\cos^2\theta=1-\left(\frac5{13}\right)^2=\frac{144}{169}.

The horizontal coordinate is negative in this quadrant, so take the negative root.

Answer. cos⁡θ=−1213\cos\theta=-\frac{12}{13}.

Check. Reflecting the first example's unit-circle point in the vertical axis preserves sine and reverses cosine. Also, (−1213)2+(513)2=1\left(-\frac{12}{13}\right)^2+\left(\frac5{13}\right)^2=1.

The identity alone cannot choose between the two signs.

Watch out

From cos⁡2θ=144169\cos^2\theta=\frac{144}{169}, writing cos⁡θ=1213\cos\theta=\frac{12}{13} automatically can lose the correct sign. Use the angle's quadrant; if no information fixes the quadrant, both signs may be possible.

Use the vertical coordinate's sign in the next calculation.

Writing tangent in sine and cosine

Tangent compares the vertical and horizontal coordinates, so its sign depends on both.

In a right triangle, dividing sin⁡θ=oppositehypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} by cos⁡θ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} cancels the hypotenuse. The quotient is oppositeadjacent=tan⁡θ\frac{\text{opposite}}{\text{adjacent}}=\tan\theta. The same signed-coordinate ratio extends this relationship to other quadrants.

Tangent identity

tan⁡θ=sin⁡θcos⁡θ,cos⁡θ≠0.\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \cos\theta\ne0.

In degrees, exclude θ=90∘+180∘k\theta=90^\circ+180^\circ k; in radians, exclude θ=π2+kπ\theta=\frac\pi2+k\pi, where kk is any integer.

On the formulae sheet.

For instance, tan⁡180∘=0−1=0\tan180^\circ=\frac0{-1}=0. At 90∘90^\circ, the denominator is zero, so tangent is undefined. A zero numerator is allowed; a zero denominator is not.

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