Properties of the Trigonometric Functions

Identities Based on the Pythagorean Theorem

Some of the most fundamental trigonometric identities are those derived from the Pythagorean Theorem. These are defined using a right triangle:

right triangle

At this stage in the course the Pythagorean Theorem should be second nature to you. If you have not yet learned it, learn it now.

$a^{2}+b^{2}=c^{2}$

$a$ and $b$ of course are the legs or the adjacent and opposite edges, and $c$ is the hypotenuse, the longest side, the side that does not include the right angle. This formula only works for a right angle triangle. If the angle shown as a right angle in the diagram were obtuse, larger than a right angle, then $c^{2}$ would be larger than the Pythagorean sum. If the angle shown as a right angle were smaller than a right angle, then $c^{2}$ would be smaller than $a^{2}+b^{2}$ .

Pythagorean Theorem in Terms of sine and cosine

The Pythagorean Theorem is the same thing as

$\sin ^{2}(A)+\cos ^{2}(A)=1$

We know this from the earlier in the book where we introduced ${\rm {sine}}$ and ${\rm {cosine}}$ .

• The ${\rm {cosine}}$ is the adjacent side when the hypotenuse is one.
• The ${\rm {sine}}$ is the opposite side when the hypotenuse is one.

The $\sin ^{2}(A)+\cos ^{2}(A)=1$ identity should be second nature to you too, but you should also be able to derive it from the Pythagorean relation.

We can see it's true from a right triangle that has a hypotenuse that is 'c' rather than one. Dividing the Pythagorean relation through by $c^{2}$ gives us

$\left({\frac {a}{c}}\right)^{2}+\left({\frac {b}{c}}\right)^{2}=\left({\frac {c}{c}}\right)^{2}=1$

And we have [[../Right angle trigonometry|already seen]] from soh-cah-toa that the sine of A is $\left({\frac {a}{c}}\right)$ and the cosine of A is $\left({\frac {b}{c}}\right)$ .

Identities based on Addition Formula

We saw the addition formula earlier.

Letters are Arbitrary

The letters in these equations $A$ and $B\,$ are arbitrary. We could equally use $\omega$ and $\theta$ . We're pointing this out because we happen to have a diagram on this page that has angles $A$ , $B$ and $C$ in it, and we want to make it clear that that was for Pythagoras and we're not talking about that diagram any more. In these equations you can also always substitute actual values, e.g. replace $A$ by $\pi$ and they'd still be true. Or you could replace $A$ by $2t$ or $\pi -t$and they'd still be true.

Addition Formula

The addition formula for cosine is:

$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$

and is worth learning. But don't learn the next one:

'Subtraction' Formula

$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$

Learning the second formula is extra effort that does not really gain you anything. You can get it instantly by replacing $B$ by $-B$ in the first equation. The only things on the right hand side that will change are the terms that had $B$ in them. The $\cos(B)$ is unchanged because $\cos(-B)=\cos(B)$ . The minus sign before the second term changes because $\sin(-B)=-\sin(B)$ .

Checking the 'subtraction' formula

Now do a quick check. If $A=B$ then $cos(A-B)$ is $\cos(0)$ which is one, right? And on the right hand side we have $\cos ^{2}(A)+\sin ^{2}(A)$ , which is also one. Looks good.

It is much better to remember it this way, because you are doing less rote learning and you're also becoming more fluent with the algebra.

Double Angle Formula

You can immediately get to the next formula by putting B equal to A in the addition formula. This is a double angle formula.

$\cos(2A)=\cos ^{2}(A)-\sin ^{2}(A)$

Half Angle Formula

Replace A by ${\frac {A}{2}}$ in the previous result and you get the half angle formula.

$\cos(A)=\cos ^{2}{\bigl (}{\tfrac {A}{2}}{\bigr )}-\sin ^{2}{\bigl (}{\tfrac {A}{2}}{\bigr )}$

Cos^2A=\2cos^2A-1

Sum of two shifted Cosines

For the next formula, you could just plough through the algebra...

$\cos(x+\theta )+\cos(x-\theta )=\dots$

Waves in and out of phase

Before we do, notice that we are adding two cosine waves, the first shifted left and the second shifted right. We've seen this before when we looked at waves in and out of phase. We're going to get another sinusoidal wave. The expression is fairly symmetric, and actually we can quickly show that the expression gives an even function, that is that the value for $-x$ is the same as for $x$ :

$\cos(-x+\theta )+\cos(-x-\theta )=$ using $\cos(x)=\cos(-x)$ we get to:

$\cos(x-\theta )+\cos(x+\theta )=$ and now swapping the two terms:

$\cos(x+\theta )+\cos(x-\theta )$ which is the original expression.

with practice you'll be able to see those steps and that the expression is an even function immediately.

A sinusoidal wave that is an even function - well it is something based on cosine. We're expecting the formula will simplify to give us something like:

$A\cos(x)$

Where $A$ will depend on $\theta$.

Values to Check With

When $\theta =0^{\circ }$ (try it in the original formula) we expect A to be 2. When $\theta =90^{\circ }$ we expect A to be 0, as the two cosine waves are $180^{\circ }$ out of phase. So now to plough through the algebra:

$\cos(x+\theta )+\cos(x-\theta )=$

$\cos(x)\cos(\theta )-\sin(x)\sin(\theta )+\cos(x)\cos(-\theta )-\sin(x)\sin(-\theta )=$

$\cos(x)\cos(\theta )-\sin(x)\sin(\theta )+\cos(x)\cos(\theta )+\sin(x)\sin(\theta )=$

$\cos(x)\cos(\theta )+\cos(x)\cos(\theta )=$

$2\cos(\theta )\cos(x)$

And to check we try with $\theta =0^{\circ }$ and get $2\cos(x)$ , and we check with $\theta =90^{\circ }$ and get $0\times \cos(x)$ , as we hoped.

Knowing roughly what we expected made it easier to get the algebraic steps right. We knew where we were going. We knew we weren't going to end up with four separate terms, some would have to cancel or combine in some other way.

Identities based on Sine and Cosine rules

The Law of Sines

${\frac {a}{\sin(A)}}={\frac {b}{\sin(B)}}={\frac {c}{\sin(C)}}$ .

This is easy to remember because it is so symmetrical. You do not even need to remember which way up the ratios are as this is also true:

${\frac {\sin(A)}{a}}={\frac {\sin(B)}{b}}={\frac {\sin(C)}{c}}$ .

What you do need to remember is how the triangle is labelled for it to be true.

The Law of Cosines

With the cosine law formula:

$a^{2}+b^{2}-2ab\cos(\theta )=c^{2}$

It is best to think of it as a more general version of Pythagoras' theorem. The a and b have to be on equal footing, so ab as a multiplier is reasonable.

If you are familiar with 'units' in physics, then use the fact that units must match up. The quantities $a,b,c$ are measurements of length. It is no good adding $a^{2}$ to $-2(a+b)\cos(\theta )$ . One might have units Km² the other units of Km. Adding $a^{2}$ to $-2ab\cos(\theta )$ the units, e.g: Km², do match.

In some more detail, it has got to be minus for $-2ab\cos(\theta )$ . That's because for $\theta <90^{\circ }$ we need $c^{2}$ to be less than for Pythagoras' theorem.

How can you remember and be sure the 2 is right? Think of an equilateral triangle with each side of length 1. The angle $\theta =60^{\circ }$ and $\cos(60^{\circ })=0.5$ . We need the 2 to get the right answer 1 + 1 + 2x1x1x0.5 = 1.

Ratio identities

$\tan(A)={\frac {\sin(A)}{\cos(A)}}$

This is one definition of tan, and is something you should just learn.

Identities based on symmetry

These symmetry identities are best remembered by remembering the graphs for these functions. Have a look at the graphs again if you don't remember them.

$0,{\frac {\pi }{2}},\pi ,{\frac {3\pi }{2}},2\pi$ are landmarks on the graphs of sine and cosine, and you should know what happens for both those functions at those landmarks. When you have that knowledge the following symmetry identities are easy to write down, because you can see them visually.

Shifting Left

$\cos(\pi +A)=-\cos(A)\quad \sin(\pi +A)=-\sin(A)\quad \tan(\pi +A)=\tan(A)$

The identities above are visualised as shifting the graph left by $\pi$ . It's a half cycle, and both cosine and sine change signs - and hence tan does not change sign, because when it is expressed as a ratio of sine over cosine both the top and bottom change signs.

Reflecting in $x={\frac {\pi }{2}}$

$\cos(\pi -A)=-\cos(A)\quad \sin(\pi -A)=\sin(A)\quad \tan(\pi -A)=-\tan(A)$

In the above we are reflecting the graph about the vertical line $x={\frac {\pi }{2}}$ . Spend enough time looking at this on the graph to see that that is true. It is important to do that to remember this trick for remembering what happens.

Now sine has a maximum at $x={\frac {\pi }{2}}$ (it has the value one) and it is symmetric about that x value, so $\sin(\pi -A)=\sin(A)$ . Cosine is zero at $x={\frac {\pi }{2}}$ . Reflecting cosine in the line $x={\frac {\pi }{2}}$ reverses the sign. $\cos(\pi -A)=-\cos(A)$

Co-function identities

$\cos(A)=\sin \left({\frac {\pi }{2}}-A\right)$

$\sin(A)=\cos \left({\frac {\pi }{2}}-A\right)$

The above two formulas are most easily seen from the right-triangle definition of cosine and sine. The two acute angles in such a triangle add up to ${\frac {\pi }{2}}$ , and we are just giving the ratios of sides lengths in terms of a different angle's trig function.

$\cos(A)=\sin \left({\frac {\pi }{2}}+A\right)$

$\sin(A)=-\cos \left({\frac {\pi }{2}}+A\right)$

These two formulae are 'the same' as the previous formulae. We get there in two steps. First subtracted pi from the angle on the right hand side, which inverts the sign, and inverted the sign to compensate. Next invert the sign of the angle on the right hand side. For cosine on the right we are done. For sine on the right we have to invert its sign since inverting the angle's sign changed the sign of the result.

Yes - keeping track of signs is tricky, and always needs care.

Resources

• Properties of the Trigonometric. Written notes created by Professor Esparza, UTSA.