Trigonometric Identities

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Applications of reciprocal and Pythagorean identities

Lesson

By keeping in mind the definitions of the various trigonometric functions together with the Pythagorean identity and the angle sum and other identities, it is often possible to simplify a complicated trigonometric expression. Typically, manipulations of this kind are needed as part of a proof of a mathematical conjecture.

The strategy will usually be to recognise that some part of an expression is identical to another expression and so can be replaced by it. When a suitable replacement is made in this way, a simplification may then be possible by collecting like terms, by the cancellation of common factors from a numerator and denominator, or by further replacements.

Simplify the expression $\left(\sin\theta-\cos\theta\right)^2-1$(`s``i``n``θ`−`c``o``s``θ`)2−1.

We begin by expanding the squared bracket. This leads to a more complicated expression but we may see that a replacement is possible. Thus, the expression becomes $\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta-1$`s``i``n`2`θ`−2`s``i``n``θ``c``o``s``θ`+`c``o``s`2`θ`−1.

We now use the identity $\sin^2\theta+\cos^2\theta\equiv1$`s``i``n`2`θ`+`c``o``s`2`θ`≡1 to make a substitution, making the expression equal to $1-2\sin\theta\cos\theta-1$1−2`s``i``n``θ``c``o``s``θ`−1. This, in turn, simplifies to $-2\sin\theta\cos\theta$−2`s``i``n``θ``c``o``s``θ`.

Finally, we use a double-angle identity to change this to $-\sin2\theta$−`s``i``n`2`θ`.

Suppose the value of $\cos\alpha$`c``o``s``α` is known to be $t$`t` and the value of $\alpha$`α` is between $\pi$π and $\frac{3\pi}{2}$3π2. What is $\tan\alpha$`t``a``n``α`?

We wish to express $\tan\alpha$`t``a``n``α` in terms of the parameter $t$`t`. By definition, we have $\tan\alpha\equiv\frac{\sin\alpha}{\cos\alpha}$`t``a``n``α`≡`s``i``n``α``c``o``s``α`. So, we begin by looking for an expression for $\sin\alpha$`s``i``n``α`.

Using $\sin^2\alpha+\cos^2\alpha\equiv1$`s``i``n`2`α`+`c``o``s`2`α`≡1, we see that $\sin\alpha\equiv\pm\sqrt{1-\cos^2\alpha}$`s``i``n``α`≡±√1−`c``o``s`2`α`. So, $$ and hence,

$$

Now, since $\alpha$`α` is between $\pi$π and $\frac{3\pi}{2}$3π2 , It must be that $\cos\alpha=t$`c``o``s``α`=`t` is negative but we know that $\tan\alpha$`t``a``n``α` is positive. Therefore, we must replace the $\pm$± with $-$− and write $$.

(Note that a square root, like $$ is always assumed to be the positive square root.)

Simplify $\tan\theta\cos^2\left(\theta\right)\sec\theta$`t``a``n``θ``c``o``s`2(`θ`)`s``e``c``θ`.

Simplify $\frac{1}{1-\sin^2\left(x\right)}-1$11−`s``i``n`2(`x`)−1.

Manipulate trigonometric expressions

Apply trigonometric methods in solving problems