Prove the identity: sin 2x / 2 cos2 x + sin x / tan x [1 + cos(90 - x)] = 1 / cos x

We are given:

To derive $\sin(A+B)$, we can replace $B$ with $-B$:

We know that $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$.
Substitute these into the equation:

5.1.2 Hence, without using a calculator, solve for $x$ in the interval $x \in [0^\circ; 360^\circ]$ if $\sqrt{3} \cos x = \sin(50^\circ+x)\cos(10^\circ-x) + \cos(50^\circ+x)\sin(10^\circ-x)$.

The RHS matches the formula for $\sin(A+B)$ derived in 5.1.1, where $A = 50^\circ+x$ and $B = 10^\circ-x$.

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.
So the equation becomes:

Divide both sides by $\sqrt{3}$:

The reference angle for $\cos x = \frac{1}{2}$ is $60^\circ$.
Cosine is positive in the first and fourth quadrants.
In the first quadrant:

In the fourth quadrant:

Both solutions are within the interval $[0^\circ; 360^\circ]$.

5.2 Prove the identity: $\frac{\sin 2x}{2 \cos^2 x} + \frac{\sin x}{\tan x [1 + \cos(90^\circ - x)]} = \frac{1}{\cos x}$.

Use the identities $\cos(90^\circ - x) = \sin x$ and $\tan x = \frac{\sin x}{\cos x}$.

Find a common denominator:

We know $\sin^2 x + \cos^2 x = 1$.

Since LHS = RHS, the identity is proven.

5.3 Show, without using a calculator, that $\sin 260^\circ = -\sqrt{\frac{\cos 20^\circ + 1}{2}}$.

Use reduction formula for $\sin 260^\circ$:

Since sine is negative in the third quadrant:

So, LHS = $-\cos 10^\circ$.

The expression $\sqrt{\frac{\cos 20^\circ + 1}{2}}$ resembles the half-angle formula for cosine.
Recall the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
Let $\theta = 10^\circ$. Then $2\theta = 20^\circ$.

Taking the square root of both sides:

(Since $10^\circ$ is in the first quadrant, $\cos 10^\circ$ is positive, so we take the positive square root).

The RHS of the original problem is $-\sqrt{\frac{\cos 20^\circ + 1}{2}}$.
Substituting the identity from Step 3:

Since LHS = $-\cos 10^\circ$ and RHS = $-\cos 10^\circ$, the identity is shown.

QUESTION 6

6.1.1 Without using a calculator, simplify the above expression to a single trigonometric ratio.

The expression is $\cos 330^\circ \cos 365^\circ + \sin 175^\circ \sin 210^\circ$.

This expression matches the compound angle formula for $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
Here, $A = 30^\circ$ and $B = 5^\circ$.

6.1.2 Hence, determine $\sin 70^\circ$ in terms of $p$ if $\cos 330^\circ \cos 365^\circ + \sin 175^\circ \sin 210^\circ = p$.

From 6.1.1, we know that $\cos 330^\circ \cos 365^\circ + \sin 175^\circ \sin 210^\circ = \cos 35^\circ$.
Given that this expression equals $p$, we have:

We need to find $\sin 70^\circ$. We can use the double angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$.
Let $\theta = 35^\circ$. Then $2\theta = 70^\circ$.

We know $\cos 35^\circ = p$. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$:

Since $35^\circ$ is in the first quadrant, $\sin 35^\circ$ is positive:

6.2 It is given that $\frac{2^{3\sin^2 x}}{4^{2\sin x}} = 1$. Determine the general solution for $x$ if $\sin x < 0$.

Rewrite the base 4 as $2^2$:

Since $1 = 2^0$, we can equate the exponents:

Factor out $\sin x$:

This gives two possible solutions for $\sin x$:

1. $\sin x = 0$
2. $3\sin x - 4 = 0 \implies 3\sin x = 4 \implies \sin x = \frac{4}{3}$

The range of the sine function is $[-1, 1]$. Therefore, $\sin x = \frac{4}{3}$ is not possible.
So, the only valid solution from the equation is $\sin x = 0$.

The problem states that we need to determine the general solution for $x$ if $\sin x < 0$.
Our only solution for $\sin x$ is $\sin x = 0$.
However, 0 is not less than 0. The condition $\sin x < 0$ is not satisfied by $\sin x = 0$.
Therefore, there is no value of $x$ that satisfies both the given equation and the condition $\sin x < 0$.