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Determine the exact values of sine, cosine, and tangent of the angle 157° 30´ then:
157° 30´ = 180° - 22° 30´ = 180° - 45° / 2
sin ( 180° - θ ) = sin θ
sin 157° 30´ = sin ( 180° - 22° 30´ ) = sin ( 180° - 45° / 2 ) = sin 45° / 2
sin 157° 30´ = sin 45° / 2
cos ( 180° - θ ) = - cos θ
cos 157° 30´ = - cos ( 180° - 22° 30´ ) = - cos ( 180° - 45° / 2 ) = - cos 45° / 2
cos 157° 30´ = - cos 45° / 2
tan ( 180° - θ ) = - tan θ
tan 157° 30´ = tan ( 180° - 22° 30´ ) = tan ( 180° - 45° / 2 ) = - tan 45° / 2
tan 157° 30´ = - tan 45° / 2
Using the half angle identitys:
sin ( x / 2 ) = ± √ [ ( 1 - cos x ) / 2 ]
sin ( 45° / 2 ) = ± √ [ ( 1 - cos 45° ) / 2 ]
In the first quadrant all trigonometric functions are positive, so:
sin ( 45° / 2 ) = √ [ ( 1 - cos 45° ) / 2 ] =
√ [ ( 1 - √ 2 / 2 ) / 2 ] = √ [ ( 2 / 2 - √ 2 / 2 ) / 2 ] =
√ [ ( 2 - √ 2 ) / 2 / 2 ] = √ [ ( 2 - √ 2 ) / 4 ] =
√ ( 2 - √ 2 ) / √ 4 = √ ( 2 - √ 2 ) / 2
cos ( x / 2 ) = ± √ [ ( 1 + cos x ) / 2 ]
cos ( 45° / 2 ) = ± √ [ ( 1 + cos 45° ) / 2 ]
In the first quadrant all trigonometric functions are positive, so:
cos ( 45° / 2 ) = √ [ ( 1 + cos 45° ) / 2 ] =
√ [ ( 1 + √ 2 / 2 ) / 2 ] = √ [ ( 2 / 2 + √ 2 / 2 ) / 2 ] =
√ [ ( 2 + √ 2 ) / 2 / 2 ] = √ [ ( 2 + √ 2 ) / 4 ] =
√ ( 2 + √ 2 ) / √ 4 = √ ( 2 + √ 2 ) / 2
tan ( x / 2 ) = ± √ [ ( 1 - cos x ) / ( 1 + cos x ) ]
tan ( 45° / 2 ) = ± √ [ ( 1 - cos 45° ) / ( 1 + cos 45° ) ]
In the first quadrant all trigonometric functions are positive, so:
tan ( 45° / 2 ) = √ [ ( 1 - cos 45° ) / ( 1 + cos 45° ) ]
tan ( 45° / 2 ) = √ [ ( 1 - cos 45° ) / ( 1 + cos 45° ) ] =
√ [ ( 1 - √ 2 / 2 ) / ( 1 + √ 2 / 2 ) ] = √ [ ( 2 / 2 - √ 2 / 2 ) / ( 2 / 2 + √ 2 / 2 ) ] =
√ [ ( 2 - √ 2 ) / 2 / ( 2 + √ 2 ) / 2 ] = √ [ ( 2 - √ 2 ) / ( 2 + √ 2 ) ] =
√ [ ( √ 2 ∙ √ 2 - √ 2 ) / ( √ 2 ∙ √ 2 + √ 2 ) ] =
√ [ √ 2 ∙ ( √ 2 - 1 ) / ( √ 2 ∙ ( √ 2 + 1 ) ] =
√ [ ( √ 2 - 1 ) / ( √ 2 + √1 ) ] =
√ [ ( √ 2 - 1 ) ∙ ( √ 2 - 1 ) / ( √ 2 + 1 ) ∙ ( √ 2 - 1 ) ] =
√ ( √ 2 - 1 )² / √ [ ( √ 2 )² - 1² ) ] =
( √ 2 - 1 ) / √ ( 2 - 1 ) =
( √ 2 - 1 ) / √ 1 = ( √ 2 - 1 ) / 1 = √ 2 - 1
So:
sin 157° 30´ = sin 45° / 2
sin 157° 30´ = √ ( 2 - √ 2 ) / 2
cos 157° 30´ = - cos 45° / 2
cos 157° 30´ = - √ ( 2 + √ 2 ) / 2
tan 157° 30´ = - tan 45° / 2
tan 157° 30´ = - ( √ 2 - 1 ) = - √ 2 + 1
tan 157° 30´ = 1 - √ 2
How do you use the half angle formulas to determine the exact values of sine, cosine, and tangent of the angle
157
∘
30
'
?
2 answers
157.5° = (1/2)315°
sin315 = -sin45 = -√2/2 and cos315 = cos45 = √2/2
using cos 2A = 1 - 2sin^2 A or 2cos^2 A - 1
cos 315 = 1 - 2sin^2 157.5°
√2/2 = 1 - 2sin^2 157.5°
sin^2 157.5° = 1 - √2/2 = (2 - √2)/4
Sin 157.5° = √ (2 - √2)/2
cos315 = 2cos^2 157.5 - 1
cos^2 157.5 = (√2/2 + 1)/2 = (√2 + 2)/4
cos 157.5 = - √(√2 + 2) / 2 , because 157.5 is in quad II
tan 157.5° = Sin 157.5°/cos 157.5°
= [√ (2 - √2)/2]/[- √(√2 + 2) / 2] = - √[ (2 - √2)/(√2 + 2) ]
which after rationalizing reduces to Bosnian's answer of 1 - √ 2
sin315 = -sin45 = -√2/2 and cos315 = cos45 = √2/2
using cos 2A = 1 - 2sin^2 A or 2cos^2 A - 1
cos 315 = 1 - 2sin^2 157.5°
√2/2 = 1 - 2sin^2 157.5°
sin^2 157.5° = 1 - √2/2 = (2 - √2)/4
Sin 157.5° = √ (2 - √2)/2
cos315 = 2cos^2 157.5 - 1
cos^2 157.5 = (√2/2 + 1)/2 = (√2 + 2)/4
cos 157.5 = - √(√2 + 2) / 2 , because 157.5 is in quad II
tan 157.5° = Sin 157.5°/cos 157.5°
= [√ (2 - √2)/2]/[- √(√2 + 2) / 2] = - √[ (2 - √2)/(√2 + 2) ]
which after rationalizing reduces to Bosnian's answer of 1 - √ 2
1 answer