2. cotŠ¤ =
3. secĀ²Š¤ - tanĀ²Š¤ = 1
4. = cosĀ²ā
5. The value of sin x is
6. cos 15Ā° = Ā
7. tan(Ī± + Ī²) =
8. sin(Ļ - Š¤) = sin Š¤
9. cos 2Š¤ =
10. sin 2Š¤ = 0.96
Step-by-step explanation:
2.
āµ cos Š¤ =
āµ 90Ā° < Š¤ < 180Ā°
- That means Š¤ lies on the 2nd quadrant
ā“ sin Š¤ is a positive value
āµ sinĀ²Š¤ + cosĀ²Š¤ = 1
ā“ sinĀ²Š¤ + ( )Ā² = 1
ā“ sinĀ²Š¤ + Ā = 1
- Subtract from both sides Ā
ā“ sinĀ²Š¤ = Ā
- Take ā for both sides
ā“ sinŠ¤ =
āµ cotŠ¤ = cosŠ¤ Ć· sinŠ¤
ā“ cotŠ¤ = Ć· Ā
ā“ cotŠ¤ =
3.
The expression is secĀ²Š¤ - tanĀ²Š¤
āµ tanĀ²Š¤ = secĀ²Š¤ - 1
- Substitute tanĀ²Š¤ by the right hand side in the expression
ā“ secĀ²Š¤ - tanĀ²Š¤ = secĀ²Š¤ - (secĀ²Š¤ - 1)
ā“ secĀ²Š¤ - tanĀ²Š¤ = secĀ²Š¤ - secĀ²Š¤ + 1
- Simplify the right hand side
ā“ secĀ²Š¤ - tanĀ²Š¤ = 1
4.
The expression is Ā
āµ The numerator is sinĀ²ā + cosĀ²ā
āµ sinĀ²ā + cosĀ²ā = 1
ā“ The numerator = 1
āµ The denominator is tanĀ²ā + 1
āµ tanĀ²ā =
ā“ tanĀ²ā + 1 = Ā + 1
- Change 1 to Ā fraction
ā“ tanĀ²ā + 1 = Ā + Ā
- Add the two fractions Ā
ā“ tanĀ²ā + 1 = Ā
āµ sinĀ²ā + cosĀ²ā = 1
ā“ tanĀ²ā + 1 =
ā“ The denominator = Ā
ā“ Ā =
- Remember denominator the denominator will be a numerator
ā“ = cosĀ²ā
5.
āµ tan x cos x =
āµ tan x =
ā“ Ć cos x =
- Simplify it by canceling cos x up with cos x down
ā“ sin x =
ā“ The value of sin x is
6.
āµ cos(Š¤ - ā) = cosŠ¤ cosā + sinŠ¤ sinā
āµ 45 - 30 = 15
ā“ cos 15Ā° = cos(45 - 30)Ā°
- Use the rule above to find the exact value
āµ cos(45 - 30)Ā° = cos 45Ā° cos 30Ā° + sin 45Ā° sin 30Ā°
āµ cos 45Ā° = and sin 45Ā° =
āµ cos 30Ā° = and sin 30Ā° =
ā“ cos(45 - 30)Ā° = Ā Ć Ā + Ā
ā“ cos(45 - 30)Ā° = Ā + Ā Ā =
ā“ cos 15Ā° = Ā
7.
āµ tan(Ī± + Ī²) =
āµ cos Ī± = and 0Ā° < Ī± < 90Ā°
- That means Ī± is in the 1st quadrant, then all trigonometry
Ā Ā ratios are positive
āµ sinĀ²Ī± + cosĀ²Ī± = 1
ā“ sinĀ²Ī± + ( )Ā² = 1
ā“ sinĀ²Ī± + Ā = 1
- Subtract Ā Ā from both sides
ā“ sinĀ²Ī± = Ā
- Take ā for both sides
ā“ sin Ī± = Ā
āµ tan Ī± = sin Ī± Ć· cos Ī±
ā“ tan Ī± = Ć·
ā“ tan Ī± = Ā
āµ sin Ī² = and 0Ā° < Ī² < 90Ā°
āµ sinĀ²Ī² + cosĀ²Ī² = 1
ā“ ( )Ā² + cosĀ²Ī² = 1
ā“ + cosĀ²Ī² = 1
- Subtract Ā Ā from both sides
ā“ cosĀ²Ī² = Ā
- Take ā for both sides
ā“ cos Ī² = Ā
āµ tan Ī² = sin Ī² Ć· cos Ī²
ā“ tan Ī² = Ć·
ā“ tan Ī² = Ā
- Substitute the values of tan Ī± and tan Ī² in the tan (Ī± + Ī²)
āµ tan(Ī± + Ī²) =
ā“ tan(Ī± + Ī²) =
ā“ tan(Ī± + Ī²) =
8.
āµ sin(Ī± - Ī²) = sin Ī± cos Ī² - cos Ī± sin Ī²
ā“ sin(Ļ - Š¤) = sin Ļ cos Š¤ - cos Ļ sin Š¤
āµ sin Ļ = 0 and cos Ļ = -1
ā“ sin(Ļ - Š¤) = (0) Ć cos Š¤ - (-1) Ć sin Š¤
ā“ sin(Ļ - Š¤) = 0 + sin Š¤
ā“ sin(Ļ - Š¤) = sin Š¤
9.
āµ cos 2Š¤ = 2 cosĀ²Š¤ - 1
āµ cos Š¤ =
ā“ cos 2Š¤ = 2 ( )Ā² - 1
ā“ cos 2Š¤ = 2 Ć - 1
ā“ cos 2Š¤ = - 1
ā“ cos 2Š¤ =
10.
āµ sin 2Š¤ = 2 sinŠ¤ cosŠ¤
āµ cosŠ¤ = 0.6 and 0Ā° < Š¤ < 90Ā°
- That means Š¤ is in the 1st quadrant and all its trigonometry
Ā Ā ratios are positive
āµ sinĀ²Š¤ + cosĀ²Š¤ = 1
ā“ sinĀ²Š¤ + (0.6)Ā² = 1
ā“ sinĀ²Š¤ + 0.36 = 1
- Subtract 0.36 from both sides
ā“ sinĀ²Š¤ = 0.64
- Take ā for both sides
ā“ sinŠ¤ = 0.8
- Substitute the values of sinŠ¤ and cosŠ¤ in the rule above
ā“ sin 2Š¤ = 2(0.8)(0.6)
ā“ sin 2Š¤ = 0.96