Verifying Trigonometric Identities

Verify that${\cot }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }={\cot }^2\theta {\cos }^2\mathrm{\theta }$ is a consistent identity.

Solution:

Step 1: Adding ${\cos }^2\mathrm{\theta }$ on both sides of the equation, we get,

$$\begin{gathered} {\cot }^2\mathrm{\theta }={\cos }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }{\cot }^2\mathrm{\theta } \\ ={\cos }^2\mathrm{\theta }(1+{\cot }^2\mathrm{\theta }) \end{gathered}$$

Step 2: It can be seen that $1+{\cot }^2\mathrm{\theta }$ is a fundamental identity, which is

Step 3: Substituting for this identity, we get

${\cot }^2\mathrm{\theta }={\cos }^2\mathrm{\theta }{\mathrm{cosec}}^2\mathrm{\theta }$

Step 4: Now, we will use a reciprocal identity for cosecant,$\mathrm{cosec\theta }=\frac{1}{\mathrm{sin\theta }}$ and squaring on both sides, we get

Step 5: Substituting the above identity, we get

${\cot }^2\mathrm{\theta }=\frac{{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}$

Step 6: We already know that $\mathrm{cot\theta }=\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}$, and after squaring it, the above step is obtained.

Hence, $LHS=RHS$. Thus the identity is verified.

Verify the identity $\frac{1+\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}=\mathrm{sin\varphi }+\mathrm{cos\varphi }$ where $\mathrm{\varphi }$ is defined over the appropriate domain.

Solution:

The LHS of the equation seems a bit complicated compared to the RHS of the equation. It is always easier to simplify the more complicated side of the equation. So let us simplify the LHS to get the RHS.

Step 1: Writing the denominator separately, we get

LHS =$\frac{1}{\mathrm{cosec\varphi }}+\frac{\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}$

Step 2: Using the reciprocal identity,

$\mathrm{sin\varphi }=\frac{1}{\mathrm{cosec\varphi }}$

we get

LHS =$\mathrm{sin\varphi }+\frac{\mathrm{cot\varphi }}{\mathrm{cosec\varphi }}$

Step 3: Using the same identity again, and $\cot \varphi =\frac{\mathrm{cos\varphi }}{\mathrm{sin\varphi }}$, we get

LHS =$\mathrm{sin\varphi }+\mathrm{cos\varphi }$

Hence we can see that $LHS=RHS$.

Thus the identity is proved.

Prove that $\mathrm{sinx}=\frac{1}{\mathrm{cosecx}}$.

Solution:

Step 1: Plot the graph of LHS, here it is sinx, and its graph looks like:

The graph of y=sinx, StudySmarter Originals

Step 2: Plot the graph of$y=\frac{1}{\cos ecx}$versus x, which looks like:

The graph of reciprocal of cosecx, StudySmarter Originals

It can be seen that the above graphs are exactly the same and hence we deduce that $\sin x=\frac{1}{\cos ecx}$.

Prove that ${\cos }^4\mathrm{\theta }-{\sin }^4\mathrm{\theta }=\cos 2\mathrm{\theta }$ is a valid trigonometric identity.

Solution:

Step 1: In this particular example, the LHS seems more complicated. Hence we will try to simplify LHS in order to arrive at RHS.

Step 2: The term${\cos }^4\mathrm{\theta }-{\sin }^4\mathrm{\theta }$ can also be written as ${\left({\cos }^2\mathrm{\theta }\right)}^2-{\left({\sin }^2\mathrm{\theta }\right)}^2$ and now it can be written as a product of two terms using the algebraic identity ${\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)$. Hence, we have:

${\left({\cos }^2\mathrm{\theta }\right)}^2-{\left({\sin }^2\mathrm{\theta }\right)}^2=\left({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }\right)\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }\right)$

Step 3: Now, we can apply the Pythagorean identity, ${\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1$,

${\cos }^4\mathrm{\theta }-{\sin }^4\mathrm{\theta }={\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }$

Step 4: Applying the Half angle formula $\cos 2\mathrm{\theta }={\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }$, we get

${\cos }^4\mathrm{\theta }-{\sin }^4\mathrm{\theta }=\cos 2\mathrm{\theta }$

which is what we are asked to prove.

Thus, $LHS=RHS$.

Is $\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=\mathrm{cosx}$ a valid trigonometric identity?

Solution:

Step 1: The LHS seems trickier than the RHS so let’s simplify LHS and check whether we arrive at RHS or not.

Step 2: It can be seen that LHS can be simplified using the algebraic identity.

$\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2$

Applying this identity respectively, we get

$\left(1+\mathrm{sinx}\right)\left(1-\mathrm{sinx}\right)=1-{\sin }^2\mathrm{x}$

Step 3: Now applying the Pythagorean identity for sine and cosine, we get

$1-{\sin }^2\mathrm{x}={\cos }^2\mathrm{x}$

which cannot be simplified further.

It can be observed that $LHS\ne RHS$ for every value of x. But for an identity to be valid, it should satisfy each value for which the function is defined.

Thus, the given trigonometric identity is false.