1) Simplify the following expressions: a) $$ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} $$ (b) $$ \frac{\sin \left(360^{\circ}- heta ight) \cos \left(180^{\circ}+ heta ight)}{\sin \left(270^{\circ} ight) \cos \left(360^{\circ}- heta ight) \sin (90+ heta)} $$ 2) Prove the following trigonometric identities. You $$ m $$ a) $$ \sin x an x+\cos x=\frac{1}{\cos x} $$ b) $$ \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 an x $$ 3) Determine the value of $$ \sin heta \cos heta $$ if $$ an heta=\frac{3}{4} $$ and Compound Angles

Question

1) Simplify the following expressions: a) $$ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} $$ (b) $$ \frac{\sin \left(360^{\circ}-	hetaight) \cos \left(180^{\circ}+	hetaight)}{\sin \left(270^{\circ}ight) \cos \left(360^{\circ}-	hetaight) \sin (90+	heta)} $$ 2) Prove the following trigonometric identities. You $$ m $$ a) $$ \sin x 	an x+\cos x=\frac{1}{\cos x} $$ b) $$ \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 	an x $$ 3) Determine the value of $$ \sin 	heta \cos 	heta $$ if $$ 	an 	heta=\frac{3}{4} $$ and Compound Angles

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\tan\left(\theta \right)=\frac{3}{4}$$
- step1: Find the domain:
  $$\tan\left(\theta \right)=\frac{3}{4},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Use the inverse trigonometric function:
  $$\theta =\arctan\left(\frac{3}{4}\right)$$
- step3: Add the period:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}$$
- step4: Check if the solution is in the defined range:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step5: Find the intersection:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}$$
Calculate or simplify the expression $$ \sin(108)^2 + \sin(18)^2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sin\left(108^{2}\right)+\sin\left(18^{2}\right)$$
- step1: Transform the expression:
  $$2\sin\left(5994\right)\cos\left(5670\right)$$
- step2: Transform the expression:
  $$\sin\left(11664\right)+\sin\left(324\right)$$
Determine whether the expression $$ \sin(x) * \tan(x) + \cos(x) = 1/\cos(x) $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\sin\left(x\right)\tan\left(x\right)+\cos\left(x\right)=\frac{1}{\cos\left(x\right)}$$
- step1: Choose a side to work on:
  $$\frac{1}{\cos\left(x\right)}=\frac{1}{\cos\left(x\right)}$$
- step2: Verify the identity:
  $$\textrm{true}$$
Solve the equation $$ \tan(\theta) = 3/4 $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\tan\left(\theta \right)=\frac{3}{4}$$
- step1: Find the domain:
  $$\tan\left(\theta \right)=\frac{3}{4},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Use the inverse trigonometric function:
  $$\theta =\arctan\left(\frac{3}{4}\right)$$
- step3: Add the period:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}$$
- step4: Check if the solution is in the defined range:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step5: Find the intersection:
  $$\theta =\arctan\left(\frac{3}{4}\right)+k\pi ,k \in \mathbb{Z}$$
Determine whether the expression $$ (\cos(x)/(1-\sin(x))) - (\cos(x)/(1+\sin(x))) = 2*\tan(x) $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\left(\frac{\cos\left(x\right)}{\left(1-\sin\left(x\right)\right)}\right)-\left(\frac{\cos\left(x\right)}{\left(1+\sin\left(x\right)\right)}\right)=2\tan\left(x\right)$$
- step1: Choose a side to work on:
  $$2\tan\left(x\right)=2\tan\left(x\right)$$
- step2: Verify the identity:
  $$\textrm{true}$$
Calculate or simplify the expression $$ (\sin(360 - \theta) * \cos(180 + \theta)) / (\sin(270) * \cos(360 - \theta) * \sin(90 + \theta)) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\left(\sin\left(360-\theta \right)\cos\left(180+\theta \right)\right)}{\left(\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)\right)}$$
- step1: Remove the parentheses:
  $$\frac{\sin\left(360-\theta \right)\cos\left(180+\theta \right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)}$$
- step2: Rewrite the expression:
  $$\frac{\sin\left(360-\theta \right)\cos\left(\theta +180\right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(90+\theta \right)}$$
- step3: Rewrite the expression:
  $$\frac{\sin\left(360-\theta \right)\cos\left(\theta +180\right)}{\sin\left(270\right)\cos\left(360-\theta \right)\sin\left(\theta +90\right)}$$
- step4: Transform the expression:
  $$\frac{\cos\left(\theta +180\right)\tan\left(360-\theta \right)}{\sin\left(270\right)\sin\left(\theta +90\right)}$$
- step5: Transform the expression:
  $$\cos\left(\theta +180\right)\tan\left(360-\theta \right)\csc\left(270\right)\csc\left(\theta +90\right)$$
- step6: Rewrite the expression:
  $$\csc\left(270\right)\cos\left(\theta +180\right)\tan\left(360-\theta \right)\csc\left(\theta +90\right)$$
Let's go through each part of your problem step by step.

### 1) Simplify the following expressions:

#### a) $$ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} $$

Using the identity $$ \sin(90^\circ + x) = \cos(x) $$, we can rewrite $$ \sin(108^\circ) $$ as:
$$
\sin(108^\circ) = \sin(90^\circ + 18^\circ) = \cos(18^\circ)
$$
Thus, we have:
$$
\sin^2(108^\circ) = \cos^2(18^\circ)
$$
Now, using the Pythagorean identity $$ \sin^2(x) + \cos^2(x) = 1 $$:
$$
\sin^2(108^\circ) + \sin^2(18^\circ) = \cos^2(18^\circ) + \sin^2(18^\circ) = 1
$$

#### b) $$ \frac{\sin \left(360^{\circ}-\theta\right) \cos \left(180^{\circ}+\theta\right)}{\sin \left(270^{\circ}\right) \cos \left(360^{\circ}-\theta\right) \sin (90+\theta)} $$

Using trigonometric identities:
- $$ \sin(360^\circ - \theta) = -\sin(\theta) $$
- $$ \cos(180^\circ + \theta) = -\cos(\theta) $$
- $$ \sin(270^\circ) = -1 $$
- $$ \cos(360^\circ - \theta) = \cos(\theta) $$
- $$ \sin(90^\circ + \theta) = \cos(\theta) $$

Substituting these into the expression gives:
$$
\frac{-\sin(\theta)(-\cos(\theta))}{-1 \cdot \cos(\theta) \cdot \cos(\theta)} = \frac{\sin(\theta) \cos(\theta)}{\cos^2(\theta)} = \tan(\theta)
$$

### 2) Prove the following trigonometric identities:

#### a) $$ \sin x \tan x+\cos x=\frac{1}{\cos x} $$

Starting with the left side:
$$
\sin x \tan x + \cos x = \sin x \cdot \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x + \cos^2 x}{\cos x} = \frac{1}{\cos x}
$$
This proves the identity.

#### b) $$ \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 \tan x $$

Finding a common denominator:
$$
\frac{\cos x(1+\sin x) - \cos x(1-\sin x)}{(1-\sin x)(1+\sin x)} = \frac{\cos x(1+\sin x - 1 + \sin x)}{1 - \sin^2 x} = \frac{2\cos x \sin x}{\cos^2 x} = 2 \tan x
$$
This proves the identity.

### 3) Determine the value of $$ \sin \theta \cos \theta $$ if $$ \tan \theta=\frac{3}{4} $$

From $$ \tan \theta = \frac{3}{4} $$, we can use the identity:
$$
\sin \theta = \frac{3}{\sqrt{3^2 + 4^2}} = \frac{3}{5}, \quad \cos \theta = \frac{4}{\sqrt{3^2 + 4^2}} = \frac{4}{5}
$$
Thus,
$$
\sin \theta \cos \theta = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}
$$

### Summary of Results:
1. a) $$ 1 $$
   b) $$ \tan(\theta) $$
2. a) Proven
   b) Proven
3. $$ \sin \theta \cos \theta = \frac{12}{25} $$
1. a) $$ 1 $$ b) $$ \tan(\theta) $$ 2. a) Proven b) Proven 3. $$ \sin \theta \cos \theta = \frac{12}{25} $$

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