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 $\left(\sin\theta-\cos\theta\right)^2-1$(sinθ−cosθ)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 $\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta-1$sin2θ−2sinθcosθ+cos2θ−1.
We now use the identity $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ≡1 to make a substitution, making the expression equal to $1-2\sin\theta\cos\theta-1$1−2sinθcosθ−1. This, in turn, simplifies to $-2\sin\theta\cos\theta$−2sinθcosθ.
Finally, we use a double-angle identity to change this to $-\sin2\theta$−sin2θ.
Suppose the value of $\cos\alpha$cosα is known to be $t$t and the value of $\alpha$α is between $\pi$π and $\frac{3\pi}{2}$3π2. What is $\tan\alpha$tanα?
We wish to express $\tan\alpha$tanα in terms of the parameter $t$t. By definition, we have $\tan\alpha\equiv\frac{\sin\alpha}{\cos\alpha}$tanα≡sinαcosα. So, we begin by looking for an expression for $\sin\alpha$sinα.
Using $\sin^2\alpha+\cos^2\alpha\equiv1$sin2α+cos2α≡1, we see that $\sin\alpha\equiv\pm\sqrt{1-\cos^2\alpha}$sinα≡±√1−cos2α. So, $$ and hence,
$$
Now, since $\alpha$α is between $\pi$π and $\frac{3\pi}{2}$3π2 , It must be that $\cos\alpha=t$cosα=t is negative but we know that $\tan\alpha$tanα 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 $\tan\theta\cos^2\left(\theta\right)\sec\theta$tanθcos2(θ)secθ.
Simplify $\frac{1}{1-\sin^2\left(x\right)}-1$11−sin2(x)−1.