F.3 MATHS Trigonometric Ratios of Special Angles 30, 45,60

2007-05-22 3:11 am
NEED STEPS
Find y in each of the following:

1) 1-2cos(y+10)=0

2) (tan2y-sin60)xtan30=cos60

3) tan y/2 x tan30 x cos30 = cos60

再加一條: 1/cos45 + 1/sin45

回答 (2)

2007-05-22 3:31 am
✔ 最佳答案
1) 1-2cos(y+10°)=0
cos(y+10°)=0.5
y+10°=60°+360°n or -60°+360°n
y=50°+360°n or -70°+360°n(where n is an integer)


2) (tan2y-sin60°)tan30°= cos60°
tan2y-sin60°=cos60°/cot60°
tan2y-sin60°=sin60°
tan2y=√3
2y=60°+180°n
y=30°+90°n(where n is an integer)

3) [tan (y/2)] (tan30°) (cos30°) = cos60°
[tan (y/2)] sin30°= cos60°
tan(y/2)=1
y/2=45°+180°n
y=90°+360°n(where n is an integer)

(4) (1/cos45 °)+ (1/sin45 °)
=(sin45°+cos45°)/(sin45°cos45°)
=√2/(1/2)
=2√2
2007-05-22 3:48 am
1) 1-2cos(y+10)=0
cos(y+10)=1/2
y+10=60
y=50

2)(tan2y-sin60)xtan30=cos60
[tan2y-(√3/2)]x(1/√3)=1/2
tan2y=√3/2+√3/2
=√3
2y=60
y=30

3)tany/2 x tan30 x cos30 = cos60
tany/2 x 1/√3 x √3/2 = 1/2
tany/2 = 1
y/2 = 45
y=90

4)1/cos45+1/sin45
=1/(1/√2) + 1/(1/√2)
=√2+√2
=2√2

√:root
I skip the degree sign..you can add it by yourself


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