Question

Please help prove these identities!

Answer 1

Hi! It will be a pleasure to help you to prove these identities, so let's get started:

PART a)

We have the following expression:

$tan(\theta)cot(\theta)-sin^{2}(\theta)=cos^2(\theta)$

We know that:

$cot(\theta)=\frac{1}{cot(\theta)}$

Therefore, by substituting in the original expression:

$tan(\theta)\left(\frac{1}{tan(\theta)}\right)-sin^{2}(\theta)=cos^2(\theta) \\ \\ \\ Simplifying: \\ \\ 1-sin^2(\theta)=cos^2(\theta)$

We know that the basic relationship between the sine and the cosine determined by the Pythagorean identity, so:

$sin^2(\theta)+cos^2(\theta)=1$

By subtracting $sin^2(\theta)$ from both sides, we get:

$\boxed{cos^2(\theta)=1-sin^2(\theta)} \ Proved!$

PART b)

We have the following expression:

$\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}=\frac{1}{1-tan(\alpha)}$

Here, let's multiply each side by $cos(\alpha)-sin(\alpha)$:

$(cos(\alpha)-sin(\alpha))\left(\frac{cos(\alpha)}{cos(\alpha)-sin(\alpha)}\right)=(cos(\alpha)-sin(\alpha))\left(\frac{1}{1-tan(\alpha)}\right) \\ \\ Then: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-tan(\alpha)}$

We also know that:

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

Then:

$cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{1-\frac{sin(\alpha)}{cos(\alpha)}} \\ \\ \\ Simplifying: \\ \\ cos(\alpha)=\frac{cos(\alpha)-sin(\alpha)}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Or: \\ \\ cos(\alpha)=\frac{\frac{cos(\alpha)-sin(\alpha)}{1}}{\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)}} \\ \\ Then: \\ \\ cos(\alpha)=cos(\alpha).\frac{cos(\alpha)-sin(\alpha)}{cos(\alpha)-sin(\alpha)} \\ \\ \boxed{cos(\alpha)=cos(\alpha)} \ Proved!$

PART c)

We have the following expression:

$\frac{cos(x+y)}{cosxsiny}=coty-tanx$

From Angle Sum Property, we know that:

$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

Substituting this in our original expression, we have:

$\frac{cos(x)cos(y)-sin(x)sin(y)}{cosxsiny}=coty-tanx$

But we can also write this as follows:

$\\ \frac{cosxcosy}{cosxsiny}-\frac{sinxsiny}{cosxsiny}=coty-tanx \\ \\ Simplifying: \\ \\ \frac{cosy}{siny}-\frac{sinx}{cosx} =coty-tanx \\ \\ But: \\ \\ \frac{cosy}{siny}=coty \\ \\ \frac{sinx}{cosx}=tanx \\ \\ Hence: \\ \\ \boxed{coty-tanx=coty-tanx} \ Proved!$

PART d)

We have the following expression:

$\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=2\ln\left|sin \theta\right|$

By Logarithm product rule, we know:

$log_{b}(x.y) = log_{b}(x) + log_{b}(y)$

So:

$\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=\ln\left|(1+cos \theta)(1-cos \theta)\right|$

The Difference of Squares states that:

$a^2-b^2=(a+b)(a-b) \\ \\ So: \\ \\ (1+cos \theta)(1-cos \theta)=1-cos^2 \theta$

Then:

$\ln\left|(1+cos \theta)(1-cos \theta)\right|=\ln\left|1-cos^{2} \theta\right|$

By the Pythagorean identity:

$sin^2(\theta)+cos^2(\theta)=1 \\ \\ So: \\ \\ sin^2 \theta = 1-cos^2 \theta$

Then:

$\ln\left|1-cos^{2} \theta\right|=\ln\left|sin^2 \theta|$

By Logarithm power rule, we know:

$log_{b}(x.y) = ylog_{b}(x)$

Then:

$\ln\left|sin^2 \theta|=2\ln\left|sin \theta|$

In conclusion:

$\boxed{\ln\left|1+cos \theta\right|+\ln\left|1-cos \theta\right|=2\ln\left|sin \theta\right|} \ Proved!$