Keep only valid sine values
Solve $2\cos^2x-3\sin x=0$ for $0\le x<2\pi$.
$$2(1-\sin^2x)-3\sin x=0$$
$$(2\sin x-1)(\sin x+2)=0.$$
$\sin x=-2$ is impossible. From $\sin x=\frac12$, the interval gives
$$x=\frac\pi6,\ \frac{5\pi}{6}.$$
Trigonometric identities and equations
Select useful identities, solve on a stated interval, and retain every valid trigonometric solution.
By the end of this lesson
- use Pythagorean, compound-angle and double-angle identities
- transform equations into a solvable form
- generate and filter all solutions on an interval
Choose an identity with a purpose
Look for a recognizable piece before rearranging. If you see $\sin^2x+\cos^2x$, replace the whole pair by 1. If an equation mixes $x$ and $2x$, a double-angle identity can put every term in the same angle.
$\sin^2x+\cos^2x=1$$\sin2x=2\sin x\cos x$$\cos2x=1-2\sin^2x=2\cos^2x-1$$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$
Choose the version that reduces the number of different functions or angles in the equation.
Checkpoint
Correct. It turns the equation into a quadratic in $\sin x$.
Not quite. Aim for one trigonometric function. The other identities retain mixed functions or angles.
Show explanation
It turns the equation into a quadratic in $\sin x$.
Example 1
Solve a quadratic form
See all the solutions
On the unit circle, cosine is horizontal position and sine is vertical position. A positive height can occur on either side of the vertical axis, so an inverse-sine value alone may miss a solution.
On the unit circle, sine is vertical height. Both π/6 and 5π/6 have height 1/2. Adding 2π repeats either point; on 0 ≤ x < 2π there are exactly these two solutions.
A full turn adds $2\pi$ without changing sine or cosine. First find the angles in one turn, then include only repeats that lie in the requested interval.
Protect solutions during algebra
Dividing by $\sin x$ or $\cos x$ can discard cases where that factor is zero. Factor first and solve each branch.
Checkpoint
Correct. A product is zero when either factor is zero.
Not quite. Dividing by $\sin x$ would lose the entire $\sin x=0$ branch.
Show explanation
A product is zero when either factor is zero.
Example 2
Use a compound angle
Exact evaluation
Find $\sin75^\circ$ exactly.
$$\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$$
$$=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12=\frac{\sqrt6+\sqrt2}{4}.$$
The signs follow directly from the addition identity.
Checkpoint
Correct. $0\le2x<2\pi$, so $2x=\pi/2,3\pi/2$.
Not quite. Transform the interval as well as the equation when the angle is multiplied.
Show explanation
$0\le2x<2\pi$, so $2x=\pi/2,3\pi/2$.
Example 3
Combine into one sinusoid
Solve $\sin x+\cos x=1$
Since $\sin x+\cos x=\sqrt2\sin(x+\pi/4)$, solve
$$\sin\left(x+\frac\pi4\right)=\frac1{\sqrt2}.$$
For $0\le x<2\pi$, this gives $x+\pi/4=\pi/4,3\pi/4$ modulo $2\pi$, hence
$$x=0,\ \frac\pi2.$$
Substitution confirms both values.
In the exam
Quick check
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Rearrange the Pythagorean identity.
Show explanation
Rearrange the Pythagorean identity.
This is the double-angle identity for sine.
Show explanation
This is the double-angle identity for sine.
Division assumes the divisor is non-zero.
Show explanation
Division assumes the divisor is non-zero.
$2\pi$ is excluded by the half-open interval.
Show explanation
$2\pi$ is excluded by the half-open interval.
$2x=2k\pi$, so $x=k\pi$.
Show explanation
$2x=2k\pi$, so $x=k\pi$.
Lesson Rewards
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