In the diagram, P(3 ; t) is a point in the Cartesian plane - NSC Mathematics - Question 5 - 2017 - Paper 2

To find tan β, we use the ratio of the opposite side to the adjacent side in the triangle formed:

tan β = rac{opposite}{adjacent} = rac{t}{3} = rac{√34}{3}

This gives:

tan β = -rac{5}{3}

Using the cosine double angle formula:

$cos 2β = 2 cos^2 β - 1$

We substitute in the value of cos β:

cos β = rac{3}{√34}

Thus, we obtain:

cos 2β = 2igg(rac{3}{√34}igg)^2 - 1

Through simplifications:

= 2 imes rac{9}{34} - 1 = rac{18}{34} - 1

This results in:

= rac{18 - 34}{34} = rac{-16}{34} = -rac{8}{17}

To prove the identity, we apply the sine addition and subtraction formulas:

$LHS = sin(A + B) - sin(A - B)$

This can be rewritten using the formula:

$= sin A cos B + cos A sin B - (sin A cos B - cos A sin B)$

Cancelling yields:

$= 2 cos A sin B$

Thus, we confirm that LHS = RHS.

Using the sine subtraction formula, we can express this as:

LHS = sin 77° - sin 43° = 2 cosigg(rac{77° + 43°}{2}igg) sinigg(rac{77° - 43°}{2}igg)

This simplifies to:

$= 2 cos(60°) sin(17°)$

Thus, since cos(60°) = rac{1}{2}, we find:

LHS = 2 imes rac{1}{2} imes sin(17°) = sin(17°)

Confirming that LHS = RHS.