In the first Quadrant, all trig. Ratios room positive. Thinking of θ together an acute angle, θ by itself ends in the 1st Quadrant at M.
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2π +θ ends in the first Quadrant in ~ M as well, specifically at the same point on the unit circle where θ ends. keep in mind that 2π means one full turn ~ above the circle.Both θand2π + θ will as such have the same values of sine, cosine, tangent, and also cotangent, as displayed below:
sin(2π + θ) = sinθ,
cos(2π + θ) = cosθ,
tan(2π + θ) = tanθ,
cot(2π + θ) = cotθ.
The Trigonometric ratios of edge 2π- θ:
Thinking that θ as an acute angle, 2π -θ ends in the 4th Quadrant where only cosine is positive. 2π -θ method going 1 complete turn native A to A and also then returning to the M" that means as much as θ. We might wirte:
Another set of crucial Formulas:
Thinking of θ together an acute angle, 2π -θ end in the 4th Quadrant where only cosine is positive. 2π -θ method going 1 full turn from A come A and then return to the M in the fourth Quadrant as much as θ. We might wirte:
sin(2π- θ) = -sinθ,
cos(2π- θ) =+cosθ,
tan(2π- θ) = -tanθ,
cot(2π- θ) = -cotθ.
The Trigonometric ratios of edge π + θ:
|Thinking the θ together an acute angle (that end in the 1st Quadrant),|
edge π+ θ end in the 3rd Quadrant where only tangent and cotangent space positive. We might write:
sin(π + θ) = -sinθ,
cos(π + θ) = -cosθ,
tan(π + θ) = +tanθ,
cot(π + θ) = +cotθ.
Connecting the M that is in the third Quadrant come O and also extending the to overcome the tangent and cotangent axes, it crosses them in their optimistic region, and also hence, both are optimistic for edge π+ θ.
The Trigonometric ratios of edge π - θ:
Thinking that θ as an acute edge (that ends in the 1st Quadrant),angle π- θ end in the second Quadrant where only sineis positive. We might write:
sin(π - θ) = +sinθ,
cos(π - θ) = -cosθ,
tan(π - θ) = -tanθ,
cot(π - θ) = -cotθ.
The Trigonometric ratios of edge π/2- θ:
Thinking ofθ as an acute edge (that ends in the 1st Quadrant), (π/2- θ) or (90�-θ) additionally ends in the1st Quadrant. due to the fact that in the first Quadrant, all trig. Ratios room positive; therefore, every trig ratios the (π/2 - θ) edge are additionally positive. What is the capture then? note that if 2 angles add up come 90�, castle are dubbed " free angles."
If two angles add up to 90�or π/2, the sine of one is same to the cosine the the other.
Also, the tangent of one is same to the cotangent the the other.θ and also (π/2-θ) are totally free angles because < θ + (π/2-θ) = π/2 >.
By introduce to the ideal triangle displayed bellow, the proofs that the above statements are:
The Trigonometric ratios of edge π/2+ θ:
Thinking ofθ as an acute angle (that ends in the 1st Quadrant), (π/2+θ) or (90�+θ) end in the2nd Quadrant where only sine of the angle is positive. The (π/2+θ) recipe are similar to the (π/2-θ) formulas except only sine is positive since (π/2+θ) ends in the 2nd Quadrant.
sin(π/2+θ) = +cosθ.
cos(π/2+θ) = -sinθ.
tan(π/2+θ) = -cotθ.
cot(π/2+θ) = -tanθ.
The Trigonometric ratios of angle(-θ):
Thinking that θ as an acute angle, (-θ)ends in the fourth Quadrant where only cosine is positive. (-θ) method starting from A, the beginning on the unit circle, and just walk clockwise (negative direction) as lot as θ.Cosine being the just positive trig proportion in the 4th Quadrant, we might write:
|sin(-θ) = -sinθ|
tan(-θ) = -tanθ
cot(-θ) = -cotθ
|same as |
|sin(2π- θ) = -sinθ.cos(2π- θ) =+cosθ.|
tan(2π- θ) = -tanθ.
cot(2π- θ) = -cotθ.
|Note that (-θ) and also (2π-θ) end at the same allude in the fourth Quadrant.|
Verify the complying with equality:
cos(80�) -2sin(10�) + 3sin(190�) - 4cot(90�) + 5sin(350�) + 6sin(170�)= -3sin(10�).
Solution: Judging by 10� and 80�, we may conclude that all angles deserve to be to express in terms of the trig. Ratios of 10�. together usual, we will certainly leave the right side untouched, and work top top the left next to check out if we acquire it equal to the appropriate side. We may write as
cos(90-10)� -2sin(10)� + 3sin(180+10)� - 4(0)� + 5sin(360-10)� + 6sin(180-10)� =
cos(π/2-10�) - 2sin(10�) + 3sin(π+10�) - 0 + 5sin(2π-10�) + 6sin(π-10�) =
+sin(10�) - 2sin(10�) + 3<-sin(10�)> -0 +5<-sin(10�)> + 6<+sin(10�)> =
This matches the appropriate side. Read THE adhering to CAREFULLY:
cos(80�) = ? Ans.: 80� end in first Quad. whereby cosine is (+); yet cos(π/2 - 10�) = sin(10�) because 80� +10� = 90� (complimentary angles)
-2sin(10�) = ?Ans.: We will leave this as it is because we re trying to transform all terms right into sin(10�) if possible; thus, -2sin(10�) = -2sin(10�).
+3sin(190�) = ?Ans.: 190� ends in the3rd Quad. Whereby sine is(-); thus, 3sin(π+10�) = -3sin10�.
-4cot(90�) = ? Ans.: -4(0) = 0.
5sin(350�) = ? Ans.: 350� end in the 4th Quad. Wheresine is (-);thus, 5sin(350�) = 5sin(2π-10�) = -5sin(10�).
6sin(170�) = ? Ans.: 170� end in the second Quad. Wherein sine is (+); thus, 6sin(170�) = 6sin(π-10�) = +6sin(10�).
Use the same layout reasoning in the verification of the complying with exercises:
Without making use of a calculator, verify the complying with equalities: (Note: Work the right side only).
1) sin (200�) + 2sin(160�) - cos(70�) + 3sin(340�) - 4cos(110�) = sin(20�).
2) 3tan(130�) + cot(40�) - tan(230�) - 5tan(310�) -2cot(140�)= 4tan(50�).
3) 2cos(75�)cos(80�) + 3cos(105�)sin(170�) + 4sin(165�)sin(190�) = -5sin(15�)sin(10�).
4) cos(115�) -3 sin(25�) - 5sin(205�) = cos(65�).
5) 3cos(310�) + cos(-50�) = 4cos(50�).
6) 2tan(60�) + 3tan(240�) = 5cot(30�).
7) sin(π + θ) + 2cos(π/2 - θ) -3sin(π - θ) = -2sin(θ).
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8) 3tan(2π - θ) + tan(π + θ) + cot(π/2 - θ) - cot(π/2 + θ) = 0.