A = sin ⁡ − 1 3 5; B = cos ⁡ − 1 5 13 A=\sin^{-1}\dfrac{3}{5}; B=\cos^{-1}\dfrac{5}{13} A = sin − 1 5 3 ; B = cos − 1 13 5 We know that for inverse sine and cosine function: sin ⁡ A = 3 5 \sin A=\dfrac{3}{5} sin A = 5 3 cos ⁡ B = 5 13 \cos B=\dfrac{5}{13} cos B = 13 5 and from page 22 we can write: sin ⁡ A = y r = 3 5 \sin A
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the followingsin A − BGiven \[ \sin A = \frac{4}{5}\text{ and }\cos B = \frac{5}{13}\]We know that\[ \cos A = \sqrt{1 - \sin^2 A}\text{ and }\sin B = \sqrt{1 - \cos^2 B} ,\text{ where }0 < A , B < \frac{\pi}{2}\]\[ \Rightarrow \cos A = \sqrt{1 - \left \frac{4}{5} \right^2} \text{ and }\sin B = \sqrt{1 - \left \frac{5}{13} \right^2}\]\[ \Rightarrow \cos A = \sqrt{1 - \frac{16}{25}}\text{ and }\sin B = \sqrt{1 - \frac{25}{169}}\]\[ \Rightarrow \cos A = \sqrt{\frac{9}{25}}\text{ and }\sin B = \sqrt{\frac{144}{169}}\]\[ \Rightarrow \cos A = \frac{3}{5}\text{ and }\sin B = \frac{12}{13}\]Now,\[\sin\left A - B \right = \sin A \cos B - \cos A \sin B \]\[ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} - \frac{36}{65}\]\[ = \frac{- 16}{65}\]
Solution The correct option is B 56 33 Explanation for the correct option: Step 1. Find the value of tan 2 α: Given, cos ( α + β) = 4 5 ⇒ sin ( α + β) = 3 5 sin ( α - β) = 5 13 ⇒ cos ( α - β) = 12 13 Now, we can write 2 α = α + β + α - β Step 2. Take " tan " on both sides, we get tan 2 α = tan ( α + β + α - β) $\begingroup$I've used the angle sum identity to end up with $\cos A \cos B -\sin A \sin B = \frac{5}{13} = \frac{3}{5}\cos B -\frac{4}{5} \sin B$, but don't know how to proceed from here. Any tips? asked Aug 10, 2017 at 1020 $\endgroup$ 1 $\begingroup$Hint $\cos B=\cosA+B-A$ use the compound angle formula answered Aug 10, 2017 at 1029 David QuinnDavid gold badges19 silver badges48 bronze badges $\endgroup$ $\begingroup$ $$A=\arcsin\dfrac45=\arccos\dfrac35$$ $$A+B=\arccos\dfrac5{13}=\arcsin\dfrac{12}{13}$$ $$B=\arccos\dfrac5{13}-\arccos\dfrac35$$ answered Aug 10, 2017 at 1056 $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .
If cos(a+b)=4/5, sin(a-b)=5/13 and 0
Given as sin A = 4/5 and cos B = 5/13 As we know that cos A = √1 – sin2 A and sin B = √1 – cos2 B, where 0 < A, B < π/2 Therefore let us find the value of sin A and cos B cos A = √1 – sin2 A = √1 – 4/52 = √1 – 16/25 = √25 – 16/25 = √9/25 = 3/5 sin B = √1 – cos2 B = √1 – 5/132 = √1 – 25/169 = √169 – 25/169 = √144/169 = 12/13 i sin A + B As we know that sin A +B = sin A cos B + cos A sin B Therefore, sin A + B = sin A cos B + cos A sin B = 4/5 × 5/13 + 3/5 × 12/13 = 20/65 + 36/65 = 20 + 36/65 = 56/65 ii cos A + B As we know that cos A +B = cos A cos B – sin A sin B Therefore, cos A + B = cos A cos B – sin A sin B = 3/5 × 5/13 – 4/5 × 12/13 = 15/65 – 48/65 = -33/65 iii sin A – B As we know that sin A – B = sin A cos B – cos A sin B Therefore, sin A – B = sin A cos B – cos A sin B = 4/5 × 5/13 – 3/5 × 12/13 = 20/65 – 36/65 = -16/65 iv cos A – B As we know that cos A - B = cos A cos B + sin A sin B Therefore, cos A - B = cos A cos B + sin A sin B = 3/5 × 5/13 + 4/5 × 12/13 = 15/65 + 48/65 = 63/65
Findthe exact value of cos (A - B) if sin A = -4/5 and cos B = 12/13 with A in quadrant III and B in quadrant IV; Question: Find the exact value of cos (A - B) if sin A = -4/5 and cos B = 12/13 with A in quadrant III and B in quadrant IV The correct option is D-1665Explanation for the correct 1 Find the value of cosA,sinBGiven that, sinA=45and cosB= know that, sin2θ+cos2θ=1cosA=1-sin2A=1-452=35Now the value of sinBis negative because B lies in 3rd quadrant. sinB=1-12132=1-144169=25169=-513Step 2 Find the value of cosA+BWe know that, cosA+B= option D is correct. 240\cos(10^{6}t-y)-240\cos10^{6}t-4.8(10^{-3})\cos(10^{6}t-y) \simeq 240\cos(10^{6}t-y)-240\cos10^{6}t (9.1.2) because the motional emf is negligible compared with the transformer emf. Using trigonometric identity, we write \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

aとb を正の実数とする．y = acosx 0 5 x 5 ˇ 2) のグラフをC1， y = bsinx 0 5 x 5 ˇ 2) のグラフをC2 とし，C1 とC2 の交点をP とする． (1) P のx 座標をt とする．このとき，sint およびcost をa とb で表せ． (2) C1; C2 とy 軸で囲まれた領域の面積S をa とb で表せ． (3) C1; C2 と直線x = ˇ 2 で囲まれた領域の面積をT

Giventhat \(\displaystyle{\sin{{A}}}={\frac{{{4}}}{{{5}}}},\ {\cos{{\left({A}+{B}\right)}}}={\frac{{{5}}}{{{13}}}}\) and that A
Example Find sin 75°, which is sin 5π/12. Solution: 75° is half of 150°, and you know the functions of 150° exactly because they are the same as the functions of 30°, give or take a minus sign. sin 75° = sin(150°/2) = ±√ (1 − cos 150°)/2. Here, cos 150° is negative because 150° is to the left of the origin, in Quadrant II, and 180° − 150° = 30°, so Hai Labibatun, jawaban soal ini adalah tan (a - b) = 33/56 Perbandingan trigonometri : • sin x = sisi depan/sisi miring • cos x = sisi samping/sisi miring • tan x = sisi depan/sisi samping » sin a = 4/5 ---> sisi depan = 4, sisi miring = 5 • sisi samping = √ ( (sisi miring)² - (sisi depan)²) sisi samping = √ (5² - 4²) sisi R8Mi.
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