5.1 Without using a calculator, simplify the following expression to ONE trigonometric ratio: sin 140° · sin(360° - x) cos 50° · tan(-x) 5.2 Prove the identity: -2sin²x + cos x + 1 1 - cos(540° - x) = 2 cos x - 1 5.3 Given: sin 36° = √(1 - p²) Without using a calculator, determine EACH of the following in terms of p: 5.3.1 tan 36° 5.3.2 cos 108° - NSC Mathematics - Question 5 - 2021 - Paper 2

To prove the identity:

$-2 ext{sin}^2x + ext{cos}x + 1 = 2 ext{cos}x - 1$

Start with the left-hand side (LHS):

$ext{LHS} = -2 ext{sin}^2x + ext{cos}x + 1$

Substituting $ext{sin}^2x = 1 - ext{cos}^2x$ into the LHS:

$ext{LHS} = -2(1 - ext{cos}^2x) + ext{cos}x + 1$

This simplifies to:

$= -2 + 2 ext{cos}^2x + ext{cos}x + 1$

Combining like terms leads to:

$= 2 ext{cos}^2x + ext{cos}x - 1$

Now, setting the LHS equal to the right-hand side (RHS):

The RHS is given as:

$ext{RHS} = 2 ext{cos}x - 1$

Thus, both sides equal:

$2 ext{cos}^2x + ext{cos}x - 1 = 2 ext{cos}x - 1$

Therefore, the identity holds:

$ext{LHS} = ext{RHS}$

Conclusion:

$ext{Identity proved!}$

Given:

ext{sin } 36° = rac{ ext{sqrt}(1-p^2)}{p}

Using the identity for tangent:

ext{tan } x = rac{ ext{sin } x}{ ext{cos } x

With the right triangle, place:

$ext{cos } 36° = ext{sqrt}(1 - ext{sin}^2 36°) = ext{sqrt}(1 - (1 - p^2)) = p$

Thus,

ext{tan } 36° = rac{ ext{sin } 36°}{ ext{cos } 36°} = rac{rac{ ext{sqrt}(1 - p^2)}{p}}{p} = rac{ ext{sqrt}(1-p^2)}{p^2}

Hence,

ext{Answer: } an 36° = rac{ ext{sqrt}(1-p^2)}{p^2}

To find cos 108°, use:

$ext{cos } 108° = ext{cos }(90° + 18°) = - ext{sin } 18°$

Using angle relationships:

$ext{cos } 108° = ext{cos }(180° - 72°) = - ext{cos } 72°$

Thus,

To express it in terms of p:

$ext{cos } 72° = ext{sin } 18° = ext{sin }(90° - 72°) = p$

Therefore,

$ext{cos } 108° = -p$

Finally,

$ext{Answer: } ext{cos } 108° = -p$