P2 6.1 - Using trigonometric identities

P2 6.1 - Using trigonometric identities

Connect sine, cosine and tangent so that an expression involving several trigonometric functions can be written in a simpler form. You will use signed values in different quadrants and prove identities by transforming one side into the other.

Sine cosine and tangent relation

If sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=−4/5\cos\theta=-4/5, can you find tan⁡θ\tan\theta without finding the angle? The signs place the angle in the second quadrant, but the link between the three functions gives the value directly.

For an acute angle in a right-angled triangle,

sin⁡θcos⁡θ=opposite/hypotenuseadjacent/hypotenuse=oppositeadjacent=tan⁡θ.\frac{\sin\theta}{\cos\theta} =\frac{\text{opposite}/\text{hypotenuse}} {\text{adjacent}/\text{hypotenuse}} =\frac{\text{opposite}}{\text{adjacent}} =\tan\theta.

The common hypotenuse cancels. The relation extends to other angles using signed coordinates: sine supplies the vertical coordinate and cosine the horizontal coordinate on a circle of radius 11.

Sine cosine and tangent relation

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

Tangent is undefined when cos⁡θ=0\cos\theta=0, for example at 90∘90^\circ and 270∘270^\circ.

Not in the formula booklet: learn it.

Signs therefore matter. In quadrant II, sine is positive and cosine negative, so tangent is negative. In quadrant III, both are negative, so tangent is positive.

Worked example 1: positive sine and cosine

Question. Given sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=4/5\cos\theta=4/5, find tan⁡θ\tan\theta.

Plan. Divide sine by cosine, cancelling the common denominator.

tan⁡θ=3/54/5=35×54=34.\tan\theta=\frac{3/5}{4/5} =\frac35\times\frac54 =\frac34.

Answer. tan⁡θ=3/4\tan\theta=3/4.

Check. (3/4)(4/5)=3/5(3/4)(4/5)=3/5, so tan⁡θcos⁡θ=sin⁡θ\tan\theta\cos\theta=\sin\theta.

Change only the sign of cosine. This is the opening question, and it changes the sign of tangent.

Worked example 2: a negative horizontal coordinate

Question. Given sin⁡θ=3/5\sin\theta=3/5 and cos⁡θ=−4/5\cos\theta=-4/5, find tan⁡θ\tan\theta.

Plan. Keep the minus sign when dividing by cosine.

tan⁡θ=3/5−4/5=35×(−54)=−34.\tan\theta=\frac{3/5}{-4/5} =\frac35\times\left(-\frac54\right) =-\frac34.

Answer. tan⁡θ=−3/4\tan\theta=-3/4.

Check. (−3/4)(−4/5)=3/5(-3/4)(-4/5)=3/5. The negative tangent also agrees with quadrant II.

Division has an order: cosine divided by sine is generally a different value.

Pythagorean identity

On a circle of radius 11, a point at angle θ\theta has coordinates (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Its squared horizontal and vertical coordinates add to the squared radius.

[DIAGRAM: asset_name: Unit circle identities; asset_slug: unit-circle-identities; caption: Signed coordinates give sine and cosine; the radius gives their squared identity.; description: A unit circle has a radius to a point in quadrant two with coordinates minus four fifths and three fifths. Horizontal and vertical projections show cosine theta equals minus four fifths and sine theta equals three fifths. The radius is one, and the labels connect the coordinate squares to one and their quotient to tangent minus three quarters.]
Signed coordinates give sine and cosine

Pythagoras gives (cos⁡θ)2+(sin⁡θ)2=12(\cos\theta)^2+(\sin\theta)^2=1^2. Squaring removes the signs, so the result holds in every quadrant, including points on the axes.

Pythagorean identity

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).

Useful rearrangements are sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta and cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta.

Not in the formula booklet: learn it.

The identity finds a missing square. To find the unsquared value, use the quadrant to decide its sign. Writing only the positive square root can lose a valid negative value.

Worked example 3: choosing a positive root

Question. Given cos⁡θ=5/13\cos\theta=5/13 and 0∘<θ<90∘0^\circ<\theta<90^\circ, find exact values of sin⁡θ\sin\theta and tan⁡θ\tan\theta.

Plan. Find sin⁡2θ\sin^2\theta, choose the sign from quadrant I, then divide sine by cosine.

sin⁡2θ=1−(513)2=144169sin⁡θ=1213positive in quadrant Itan⁡θ=12/135/13=125\begin{aligned} \sin^2\theta&=1-\left(\frac5{13}\right)^2 =\frac{144}{169}\\ \sin\theta&=\frac{12}{13} &&\text{positive in quadrant I}\\ \tan\theta&=\frac{12/13}{5/13}=\frac{12}{5} \end{aligned}

Answer. sin⁡θ=12/13\sin\theta=12/13 and tan⁡θ=12/5\tan\theta=12/5.

Check. 122+52=13212^2+5^2=13^2, and the three values are positive as expected for an acute angle.

Change only the stated quadrant. The supplied cosine is unchanged, but sine and tangent must now be negative.

Worked example 4: choosing a negative root

Question. Given cos⁡θ=5/13\cos\theta=5/13 and 270∘<θ<360∘270^\circ<\theta<360^\circ, find exact values of sin⁡θ\sin\theta and tan⁡θ\tan\theta.

Plan. Use the same squared identity, then choose the quadrant IV sign before finding tangent.

sin⁡2θ=1−25169=144169sin⁡θ=−1213negative in quadrant IVtan⁡θ=−12/135/13=−125\begin{aligned} \sin^2\theta&=1-\frac{25}{169}=\frac{144}{169}\\ \sin\theta&=-\frac{12}{13} &&\text{negative in quadrant IV}\\ \tan\theta&=\frac{-12/13}{5/13}=-\frac{12}{5} \end{aligned}

Answer. sin⁡θ=−12/13\sin\theta=-12/13 and tan⁡θ=−12/5\tan\theta=-12/5.

Check. (−12/13)2+(5/13)2=1(-12/13)^2+(5/13)^2=1, while negative divided by positive gives a negative tangent.

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