Radians and Trigonometric Identities

Radians #

The angle whose arc equals the radius.

A radian is the angle whose arc is exactly as long as the radius

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Lay a length equal to the radius along the circumference. The angle that arc makes at the center is one radian.

The arc it cuts is exactly as long as the radius — that is the definition of a radian.

Half the circumference is $\pi r$πr, and each radian covers r of it — so $\pi$π radians fit.

It takes $\pi$π radians to make half a turn: $180^{\circ }=\pi$180° = π radians.

Measured this way, arc length is radius × angle: $s=r\theta$s = rθ.

Exact Trigonometric Values in Radians #

The same nine values, under their radian names.

The nine exact ratios keep their values when the angle is written in radians

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In radians, 30° is $\frac{\pi }{6}$π/6, 45° is $\frac{\pi }{4}$π/4, and 60° is $\frac{\pi }{3}$π/3.

Nothing about the shape changed, so $\sin \left(\frac{\pi }{4}\right)$sin(π/4) is the same $\frac{1}{\sqrt{2}}$1/√2 as sin 45°.

These are the same nine values under their radian names — exams write angles this way.

Two more worth knowing: $\sin \left(\frac{\pi }{2}\right)=1$sin(π/2) = 1 and $\cos \left(\frac{\pi }{2}\right)=0$cos(π/2) = 0, at the quarter turn.

sin 0 = 0 and cos 0 = 1; past $\frac{\pi }{2}$π/2 the same sizes return, but cosine and tangent become negative.

Arc Length and Sector Area #

Where the radian earns its keep.

In radians, arc length is $r\theta$rθ and sector area is half of $r^2$r² times $\theta$θ

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From $s=r\theta$s = rθ: 2 radians on radius 6 gives an arc of length 12.

A sector is the slice between two radii. Its fraction of the whole circle is $\theta$θ out of $2\pi$2π.

Take that fraction of $\pi r^2$πr² and cancel the $\pi$π: sector area $=\frac{1}{2}{r}^2\theta$= ½r²θ.

With r = 6 and $\theta =2$θ = 2, the area is ½ · 36 · 2 = 36 — no $\pi$π appears.

Trigonometric Identities #

Pythagoras, read off the unit circle.

Pythagoras on the unit circle gives an identity true for every angle

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Draw a circle of radius 1 and take any point on its circumference.

Let the radius to that point make an angle $\theta$θ with the horizontal. Its horizontal and vertical distances from the center are then $\cos \theta$cos θ and $\sin \theta$sin θ.

Pythagoras on that triangle: ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1.

Divide them and you get the tangent: $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ.

Proving a Trigonometric Identity #

Work one side; never cross the equals sign.

Walk one side to the other with known identities — never move terms across the equals sign

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Work on one side only: write $\tan \theta +\cot \theta$tan θ + cot θ in sines and cosines, then replace ${\sin }^2\theta +{\cos }^2\theta$sin²θ + cos²θ with 1.

A proof walks one side to the other. Moving terms across the equals sign assumes what you are proving.

Prove $\left(1-\cos \theta \right)\left(1+\cos \theta \right)={\sin }^2\theta$(1 − cos θ)(1 + cos θ) = sin²θ: expand, then swap $1-{\cos }^2\theta$1 − cos²θ for ${\sin }^2\theta$sin²θ.

The Compound Angle Formulas #

sin(A+B) refuses to split naively.

sin(A+B) is not sin A + sin B — each sine pairs with the other angle’s cosine

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Suppose sin(A+B) = sin A + sin B. Test it at 45° + 45°: the two sides disagree.

The correct rule pairs each sine with the other angle’s cosine: sin A cos B + cos A sin B.

Test it with A = 30° and B = 60°: it gives exactly sin 90° = 1.

The cosine rule pairs cosine with cosine and sine with sine, and the middle sign is a minus.

The proof stacks two right triangles inside the angle A + B. Every check above agrees with the formula.

The Double Angle Formulas #

One substitution doubles the angle.

Set B equal to A and the compound formulas collapse into the double angles

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Put B = A in the sine formula: sin 2A = 2 sin A cos A.

The same move in cosine: $\cos 2A={\cos }^{2}A-{\sin }^{2}A$cos 2A = cos²A − sin²A.

Replace ${\cos }^{2}A$cos²A with $1-{\sin }^{2}A$1 − sin²A, and cos 2A becomes $1-2{\sin }^{2}A$1 − 2sin²A.

Rearranged, it gives ${\sin }^{2}A$sin²A on its own — the form needed to integrate ${\sin }^{2}A$sin²A.

Secant, Cosecant and Cotangent #

The three ratios, turned upside down.

Each trig ratio has a reciprocal, and the names of the pairs cross over

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Cosine has a reciprocal called the secant: $\sec \theta$sec θ is 1 over $\cos \theta$cos θ.

Two more reciprocals complete the set: $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$cosec θ = 1/sin θ and $\cot \theta =\frac{1}{\tan \theta }$cot θ = 1/tan θ.

The names cross over: secant is the reciprocal of cosine, and cosecant of sine.

Any value you already know gives its reciprocal at once: cos 60° is ½, so sec 60° is 2.

The Identity 1 + tan²θ = sec²θ #

Pythagoras, divided through by ${\cos }^2$cos².

Divide ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1 by ${\cos }^2\theta$cos²θ and it becomes $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ

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Start from the identity already proved: ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1, true at every angle.

Divide every term by ${\cos }^2\theta$cos²θ. The middle term collapses to 1.

Write $\frac{{\sin }^2\theta }{{\cos }^2\theta }$sin²θ/cos²θ as ${\tan }^2\theta$tan²θ and $\frac{1}{{\cos }^2\theta }$1/cos²θ as ${\sec }^2\theta$sec²θ: $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ.

Dividing instead by ${\sin }^2\theta$sin²θ gives the matching identity: $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$1 + cot²θ = cosec²θ.

The Harmonic Form R sin(x + α) #

Two waves collapse into one.

a sin x + b cos x is a single wave, $R\sin \left(x+\alpha \right)$R sin(x + α), with $R=\sqrt{a^2+b^2}$R = √(a² + b²)

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Expand $R\sin \left(x+\alpha \right)$R sin(x + α) with the compound formula. It has the same shape as a sin x + b cos x.

Match coefficients: a must be $R\cos \alpha$R cos α and b must be $R\sin \alpha$R sin α.

Square both and add: $a^2+b^2=R^2$a² + b² = R², because ${\sin }^2\alpha +{\cos }^2\alpha =1$sin²α + cos²α = 1. So $R=\sqrt{a^2+b^2}$R = √(a² + b²).

So 3 sin x + 4 cos x is exactly $5\sin \left(x+\alpha \right)$5 sin(x + α), with $\tan \alpha =\frac{4}{3}$tan α = 4/3.

It is a single wave with amplitude R, so 3 sin x + 4 cos x never rises above 5.

Solving a sin x + b cos x = c with the R Form #

One wave to solve, and its crest priced at R.

Once a sin x + b cos x is one wave, an equation in it is simple and its extremes are R and −R

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Write the left side as one wave, then divide by R. A simple equation is left.

The bracket carries the range with it, so x + 53.1° starts at 53.1°, not at 0°.

Write u for x + 53.1°. Three angles near the range have sine 0.5, and 30° falls below 53.1°, so it is dropped.

Subtract $\alpha$α from each angle that is left: x = 96.9° and x = 336.9°.

The wave has amplitude 5, and the line at 2.5 cuts it twice — at the two answers just found.

The same form gives the extremes: the greatest value is R and the least is −R.

Every Solution of a Trigonometric Equation #

Every lap of the circle repeats the answers.

Trig equations repeat their answers every full turn — one formula lists them all

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sin x = ½ at 30° and at 150°, and again after every full turn of 360°.

Add any number of whole turns: 30° + 360°n and 150° + 360°n list every one.

Cosine’s first two answers are 60° and −60°, so its list is $\pm 60^{\circ }+360^{\circ }n$±60° + 360°n.

Tangent repeats every 180°, so 45° + 180°n covers every solution.