Mt5, Mt6 - Fundamental trigonometric identities

Mt5, Mt6 - Fundamental trigonometric identities

Use the tangent ratio in Mt5 and the Pythagorean identity in Mt6 to connect exact values, prove equalities and solve equations. The emphasis is on choosing a useful identity while keeping signs, denominators and all permitted solutions under control.

Two identities from coordinates

Can you find sine and cosine exactly from tan⁡θ=2\tan\theta=2, without finding the angle? A unit-circle point with that ratio lies on y=2xy=2x. Substituting into x2+y2=1x^2+y^2=1 gives 5x2=15x^2=1. There are two possible points, one in quadrant I and one in quadrant III. By the end you will choose the right one from an interval.

You already know

The point on the unit circle at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta). Pythagoras gives its squared distance from the origin; quadrant signs tell you which coordinates are positive or negative.

This coordinate picture produces two identities. One is a sum of squares; the other is a ratio of signed coordinates.

A line of slope 2 meets the unit circle in quadrants I and III

Fundamental trigonometric identities

sin⁡2θ+cos⁡2θ=1(θ∈R)\sin^2\theta+\cos^2\theta=1\qquad(\theta\in\mathbb R) tan⁡θ=sin⁡θcos⁡θ(cos⁡θ≠0)\tan\theta=\frac{\sin\theta}{\cos\theta}\qquad(\cos\theta\ne0)

Not in the formula sheet: learn it.

The first follows from x2+y2=1x^2+y^2=1. The second follows from the slope y/xy/x, so it excludes θ=π/2+kπ\theta=\pi/2+k\pi, with integer kk. At π/2\pi/2, the first identity remains true while the tangent ratio is undefined. Here sin⁡2θ=(sin⁡θ)2\sin^2\theta=(\sin\theta)^2, rather than sin⁡(θ2)\sin(\theta^2).

Identity

An identity is an equality true for every input in its stated domain. An equation selects the inputs where an equality holds. A few successful numerical checks are evidence, but cannot establish an identity for every input.

A small change can destroy an identity: at θ=π/4\theta=\pi/4, sin⁡θ+cos⁡θ=2\sin\theta+\cos\theta=\sqrt2, not 11. Yet at 00 that sum does equal 11. The squares in the real identity carry the geometry.

Key idea

Use the sum of squares to find magnitudes and the ratio to relate signed values. Neither removes the need to check the domain.

A domain error can hide in a calculation that uses an otherwise correct identity.

Recovering values without losing signs

A tangent value gives a ratio; combine it with the sum of squares and the quadrant to recover the two coordinates.

The opening problem has an unexpected feature: the tangent is an integer, but both sine and cosine are irrational. A simple ratio need not have simple rational components.

Worked example 1: a ratio in quadrant III

Question. Given tan⁡θ=2\tan\theta=2 and π<θ<3π/2\pi<\theta<3\pi/2, find sin⁡θ\sin\theta and cos⁡θ\cos\theta exactly.

Plan. Write sine in terms of cosine, substitute into the identity, then choose the negative signs.

sin⁡θ=2cos⁡θtangent ratio4cos⁡2θ+cos⁡2θ=1sum of squarescos⁡2θ=1/5collect like termscos⁡θ=−1/5,sin⁡θ=−2/5quadrant III\begin{aligned} \sin\theta&=2\cos\theta && \text{tangent ratio}\\ 4\cos^2\theta+\cos^2\theta&=1 && \text{sum of squares}\\ \cos^2\theta&=1/5 && \text{collect like terms}\\ \cos\theta&=-1/\sqrt5,\quad\sin\theta=-2/\sqrt5 && \text{quadrant III} \end{aligned}

Why this step? Tangent is positive in two quadrants; the supplied interval selects the point with two negative coordinates.

Answer. cos⁡θ=−5/5\cos\theta=-\sqrt5/5 and sin⁡θ=−25/5\sin\theta=-2\sqrt5/5.

Check. Their ratio is 22, and their squares sum to 1/5+4/5=11/5+4/5=1.

Change only the interval to quadrant I, keeping the tangent value fixed.

Worked example 2: the other point with the same ratio

Question. Given tan⁡θ=2\tan\theta=2 and 0<θ<π/20<\theta<\pi/2, find sin⁡θ\sin\theta and cos⁡θ\cos\theta exactly.

Plan. Reuse the magnitude calculation, then choose the positive coordinate signs.

sin⁡θ=2cos⁡θsame ratio5cos⁡2θ=1same squared identitycos⁡θ=1/5,sin⁡θ=2/5quadrant I\begin{aligned} \sin\theta&=2\cos\theta && \text{same ratio}\\ 5\cos^2\theta&=1 && \text{same squared identity}\\ \cos\theta&=1/\sqrt5,\quad\sin\theta=2/\sqrt5 && \text{quadrant I} \end{aligned}

Why this step? The squared equation cannot distinguish the two points; the interval does.

Answer. cos⁡θ=5/5\cos\theta=\sqrt5/5, sin⁡θ=25/5\sin\theta=2\sqrt5/5.

Check. Both values are positive, have magnitudes below 11 and give tangent 22.

Starting from sine or cosine is equally useful. If sin⁡θ=3/5\sin\theta=3/5 in quadrant II, then cos⁡2θ=16/25\cos^2\theta=16/25, giving cos⁡θ=−4/5\cos\theta=-4/5, tan⁡θ=−3/4\tan\theta=-3/4. If cos⁡θ=−5/13\cos\theta=-5/13 in quadrant III, then sin⁡θ=−12/13\sin\theta=-12/13 and tan⁡θ=12/5\tan\theta=12/5. An interval is the missing sign information in both calculations.

Watch out

Wrong: choose a positive square root simply because the square is positive. Right: the square root gives a magnitude; the quadrant supplies the sign of sine or cosine.

Change the tangent value, retaining the quadrant-III choice from the first example.

Simplifying and proving with domains

Choose a form of the identity that removes an awkward square or produces a factor you can cancel, while keeping the starting expression's restrictions.

For example,

1−cos⁡2θsin⁡θ=sin⁡θ(sin⁡θ≠0),1−sin⁡2θcos⁡θ=cos⁡θ(cos⁡θ≠0)\frac{1-\cos^2\theta}{\sin\theta}=\sin\theta\quad(\sin\theta\ne0),\qquad \frac{1-\sin^2\theta}{\cos\theta}=\cos\theta\quad(\cos\theta\ne0)

The simpler result may exist where the quotient does not. That does not fill a missing input in the quotient. A useful related proof is

1−cos⁡θsin⁡θ=1−cos⁡2θsin⁡θ(1+cos⁡θ)=sin⁡θ1+cos⁡θ(sin⁡θ≠0)\frac{1-\cos\theta}{\sin\theta} =\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)} =\frac{\sin\theta}{1+\cos\theta}\qquad(\sin\theta\ne0)

Here the conjugate 1+cos⁡θ1+\cos\theta turns the numerator into a difference of squares. The stated condition ensures all denominators and cancelled factors are nonzero.

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