Misc 10 Find sin 𝑥/2, cos 𝑥/2 and tan 𝑥/2 for sin𝑥 = 1/4 , 𝑥 in quadrant II Given that x is in quadrant II So, 90° < x < 180° Replacing x with 𝑥/2 (90°)/2 < 𝑥/2 < (180°)/2 45° < 𝑥/2 < 90° So, 𝑥/2 lies in Ist quadrant In Ist quadrant, sin , cos & tan are positive sin 𝑥/2 ,Tan a=4/3, condition pi/2 Cos b = 1/2, condition 0 Need exact value cos(ab) Sin(ab) Tan(ab) I believe a would be in quadrant 2 where x is negative and y is positiveTrigonometry Find the Other Trig Values in Quadrant I cos (s)=3/4 cos (s) = 3 4 cos ( s) = 3 4 Use the definition of cosine to find the known sides of the unit circle right triangle The quadrant determines the sign on each of the values cos(s) = adjacent hypotenuse cos (
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Quadrant 1 2 3 4 sin cos tan-The graph of cos the same as the graph of sin though it is shifted 90° to the right/ left For this reason sinx = cos (90 x) and cosx = sin (90 x) Note that cos is an even function it is symmetrical in the yaxis sin is an odd function The graph of tan has asymptotes Summary First Quadrant All are positive in this quadrant Second Quadrant Only sin is positive in this quadrant Third Quadrant Only tan is positive in this quadrant Fourth Quadrant Only cos is positive in this quadrant We now consider angles in cartesian plane We divide the plane into four quadrants in the anticlockwise sense as shown
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You can put this solution on YOUR website!Cosine cos (210°) = −1732 / 2 = −0866 Tangent tan (210°) = −1 / −1732 = 0577 Note Tangent is positive because dividing a negative by a negative gives a positive In Quadrant IV, sine and tangent are negative Since $\cos^2t\sin^2t=1$, dividing both sides by $\cos^2 t$ we also have $$1\tan^2t=\frac 1{\cos^2t}$$ Also, in the second quadrant, $\cos t0$ Use the second equation and the restriction to find $\cos t$, then use the first equation and the restriction to find $\sin t$ Then add those for your final answer
⇒ cos A = cos 2 4 0 0 ⇒ A = 2 4 0 0 Now, c o s B = − 2 1 ⇒ c o s B = − cos 6 0 0 when B does not lie in the third quadrant ⇒ c o s B = cos (1 8 0 0 − 6 0 0) ⇒ c o s B = cos 1 2 0 0 ⇒ B = 1 2 0 0 Substituting the value of A and B in equation (1) and we get, tan 1 2 0 0 sin 2 4 0 0 4 sin 1 2 0 0 − 3 tan 2 4 0 0 ⇒ tan (1When angle a is in Quadrant 3 (between 180° and 270°), both the adjacent and the opposite side are negative Hence, Sine and Cosine are negative and since Tangent (T) is a division between two negative numbers, it is the only trigonometric function that is positive Sine, Cosine and Tangent in Quadrant 4 Transcript Let s see the angles in different Quadrants In Quadrant 1, angles are from 0 to 90 In Quadrant 2, angles are from 90 to 180 In Quadrant 3, angles are from 180 to 270 In Quadrant 4, angles are from 270 to 360 To learn sign of sin, cos, tan in different quadrants, we remember Add Sugar To Coffee Representing as a table Quadrant I Quadrant II Quadrant III Quadrant IV sin cos tan
We know that sin^2 a cos^2 a= 1 ==> cos^2 a = 1 sin^2 a ==> cos^2 a = 1 (2/3)^2 ==> cos^2 a = 1 4/9 ==> cos^2 a= 5/9 ==> cos a = sqrt5/3 But a is in the 2nd quadrant where cos a iaQuestion Find the exact value, given that cos A= 1/3, with A in quadrant I, and sin B= 1/2, with B in quadrant IV, and sin C= 1/4, with C in quadrant II sin(AB) I don't understand this problem Why do they give sin C? 1 Answer Dean R tanθ = 3 4 means an opposite of 3, an adjacent of 4, so a hypotenuse of 5, because 32 42 = 52 cosθ = adjacent hypotenuse = ± 4 5
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Please help me Found 2 solutions by lwsshak3, jim_thompson5910If Tan Theta 3 4 And Theta Is Not In First Quadrant Then Sin Pi 2 Theta Cot Pi Theta YoutubeFind the Other Trig Values in Quadrant I csc (x)=4 csc(x) = 4 csc ( x) = 4 Use the definition of cosecant to find the known sides of the unit circle right triangle The quadrant determines the sign on each of the values csc(x) = hypotenuse opposite csc ( x) = hypotenuse opposite
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A) quadrant 2 or 3 b) Quadrant 2 sin , cos , tan Quadrant 3 sin , cos , tan c) 115°, 245° 13 14 Answers may vary For example, given P (x, y) on the terminal arm of angle , sin , cos , and tan 15 a) 25°, 155°, 5°, 335° b) 148°, 352 o c) 16°, 106 o, 196 o, 286 o 16 a) could lie in quadrant 3 or 4 5 233° or 307° b) could lie Now Use the formula cos(A− B) = cosAcosB sinAsinB to evaluate Note that A = tan−1( 4 3) and B = sin−1( 12 13) Therefore, cos(tan−1(4 3) − sin−1( 12 13)) = cos(tan−1(4 3))cos(sin−1( 12 13)) sin(tan−1( 4 3))sin(sin−1(12 13)) = cosAcosB sinAsinB > Use triangles A and B to find the ratios = 3 5 ⋅ 5 13 4 5 ⋅ 12 13Answer (1 of 7) Here, tan x = 1/2 We know that, the relation between tan x and sec x is sec^2xtan^2x=1 Using this relation, first, we will determine the value of cos x and then, we will determine the value of sin x So now, sec^2xtan^2x=1 => sec^2x(1/4)=1 => sec^2x=1(1/4) => sec^2x=
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