(8.5) $$ \frac{3 \cos 150^{\circ} \cdot \sin 270^{\circ}}{ an \left(-45^{\circ} ight) \cdot \cos 600^{\circ}} $$ (8.6) $$ \frac{ an \left(180^{\circ}-x ight) \cdot \sin \left(90^{\circ}+x ight)}{\sin (-x)}-\sin y \cdot \cos \left(90^{\circ}-y ight) $$ $$ 8.7 \frac{ an 30^{\circ} \cdot \sin 60^{\circ} \cdot \cos 25^{\circ}}{\cos 135^{\circ} \cdot \sin \left(-45^{\circ} ight) \cdot \sin 65^{\circ}} $$ $$ 8.8 \frac{ an \left(180^{\circ}-x ight) \cdot \sin \left(90^{\circ}-x ight)}{\cos \left(90^{\circ}+x ight)}-\frac{\cos \left(180^{\circ}-x ight)}{\sin \left(90^{\circ}+x ight)} $$ 8.9 $$ \frac{\sin 189^{\circ}}{ an 549^{\circ}}-\frac{\cos ^{2}\left(-9^{\circ} ight)}{\sin 99^{\circ}} $$ Solving trigonometric equations (no calculator 1. If $$ \sin \mathrm{A}=\frac{-3}{5} $$ and $$ 0^{\circ}

Question

(8.5) $$ \frac{3 \cos 150^{\circ} \cdot \sin 270^{\circ}}{	an \left(-45^{\circ}ight) \cdot \cos 600^{\circ}} $$ (8.6) $$ \frac{	an \left(180^{\circ}-xight) \cdot \sin \left(90^{\circ}+xight)}{\sin (-x)}-\sin y \cdot \cos \left(90^{\circ}-yight) $$ $$ 8.7 \frac{	an 30^{\circ} \cdot \sin 60^{\circ} \cdot \cos 25^{\circ}}{\cos 135^{\circ} \cdot \sin \left(-45^{\circ}ight) \cdot \sin 65^{\circ}} $$ $$ 8.8 \frac{	an \left(180^{\circ}-xight) \cdot \sin \left(90^{\circ}-xight)}{\cos \left(90^{\circ}+xight)}-\frac{\cos \left(180^{\circ}-xight)}{\sin \left(90^{\circ}+xight)} $$ 8.9 $$ \frac{\sin 189^{\circ}}{	an 549^{\circ}}-\frac{\cos ^{2}\left(-9^{\circ}ight)}{\sin 99^{\circ}} $$ Solving trigonometric equations (no calculator 1. If $$ \sin \mathrm{A}=\frac{-3}{5} $$ and $$ 0^{\circ}<\mathrm{A}<270^{\circ} $$ determine the value of: 1.1 $$ \cos \mathrm{A} $$ $$ 1.2 	an \mathrm{~A} $$. (2.) If $$ -5 	an 	heta-3=0 $$ and $$ \sin 	heta<0 $$, determine: $$ 2.1 \sin ^{2} 	heta^{\circ} $$ $$ 2.25 \cos 	heta $$ $$ 2.31-\cos ^{2} 	heta $$. 3. If $$ 13 \cos 	heta+12=0 $$ and $$ 180^{\circ}<	heta<360^{\circ} $$, evaluate: $$ 3.1 \sin 	heta \cos 	heta $$ $$ 3.2 	an 	heta $$ $$ 3.3 \sin ^{2} 	heta+\cos ^{2} 	heta $$. (4.) If $$ 3 	an 	heta-2=0 $$ and $$ 	heta \in\left[90^{\circ} ; 360^{\circ}ight] $$, determine, the value of $$ \sqrt{13}(\sin 	heta- $$

UpStudy AI · Accepted Answer

Below are one acceptable set of solutions. (Any answer equivalent to the answers below is correct.)

──────────────────────────────
Part (8): Simplify the expressions

• (8.5) Evaluate 
   (3·cos150°·sin270°)⁄(tan(–45°)·cos600°).

Step‑1. Find the standard values:
  cos150° = cos(180°–30°) = –cos30° = –(√3⁄2),
  sin270° = –1,
  tan(–45°) = –tan45° = –1,
  cos600°. Since 600°–360° = 240° and 240° = 180°+60°, 
    cos240° = –cos60° = –(1⁄2).

Step‑2. Substitute:
  Numerator = 3·(–(√3⁄2))·(–1) = 3(√3⁄2),
  Denom.  = (–1)·(–1⁄2) = 1⁄2.
  Thus, (3√3⁄2)/(1⁄2) = 3√3.

Answer (8.5): 3√3

──────────────────────────────
• (8.6) Simplify 
   [tan(180°–x)·sin(90°+x)]⁄[sin(–x)] – sin y·cos(90°–y).

Step‑1. Use identities:
  tan(180°–x) = –tan x,
  sin(90°+x) = cos x,
  sin(–x) = –sin x.
  So the first fraction becomes
   [ (–tan x·cos x) ]⁄(–sin x) = (tan x·cos x)⁄sin x.
  But tan x = sin x⁄cos x, so (sin x⁄cos x·cos x)⁄sin x = 1.

Step‑2. In the second term, note that
  cos(90°–y) = sin y, so
   sin y·cos(90°–y) = sin y·sin y = sin²y.

Thus the whole expression equals
  1 – sin²y = cos²y   (using 1 – sin²y = cos²y).

Answer (8.6): cos²y

──────────────────────────────
• (8.7) Simplify 
   [tan30°·sin60°·cos25°]⁄[cos135°·sin(–45°)·sin65°].

Step‑1. Compute the standard values:
  tan30° = 1⁄√3,
  sin60° = √3⁄2,
  so tan30°·sin60° = (1⁄√3)(√3⁄2
(8.5) 3√3 (8.6) cos²y (8.7) (1⁄√3)(√3⁄2)·cos25°⁄[cos135°·sin(–45°)·sin65°]

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Trigonometry Questions