Webb14 okt. 2024 · Introduction to Trigonometry Class 10 Extra Questions Objective Type Question 1. The value of will be: (i) sin 60° (ii) cos 60° (iii) tan 60° (iv) sin 30° Answer: (ii) cos 60° Solution. = 2 sin 30o. cos 30° = sin 60° Hence, Choice (ii) is correct Question 2. If sin θ = 1, the value of sin 20 will be : (i) -1 (ii) 0 (iii) 1 (iv) 2 Answer: (ii) 0 WebbClick here👆to get an answer to your question ️ Prove that sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B. Solve Study Textbooks Guides. Join / Login >> Class 11 >> Maths >> Trigonometric Functions >> Trigonometric Functions of Sum and Difference of Two angles

sin (a+b) * sin (a-b) = sin^2a+-sin^2b доказать тождество

Webb11 dec. 2008 · 正弦定理を使います。 ABCにおいてBC=a、CA=b、AB=cとおくと正弦定理より sinA/a=sinB/b=sinC/c よって sin^2A/a^2=sin^2B/b^2=sin^2C/c^2 sin^2A/a^2=sin^2B/b^2=sin^2C/c^2=tとおけば sin^2A=a^2t sin^2B=b^2t sin^2C=c^2t 与えられた等式sin^2A+sin^2B=sin^2Cに代入して (a^2+b^2)t=c^2t a^2+b^2=c^2 よって … trying my best anson seabra piano sheet music

若2a=3，2b=5，2c=30，试用a，b表示c． - 雨露学习互助

Webb18.在 ABC中，已知 sin^2A+sin^2B=sin^2C+sinA+sinB1）求角C; (2）若c=4，求a+b的最大值. 答案. 18.解: (1)由 sin^2A+sin^2B=sin^2C+sinA+sinB得 a^2+b^2=c^2+ab 所以 cosC= (a^2+b^2-c^2)/ (2ab)=1/2又 0Cπ ，故角 C=π/ (3) (2)因为c=4,所以 16=a^2+b^2-ab= (a+b)^2-3ab .-10分又 ab≤ ( (a+b)/2)^2 b2， 16≥1/4 (a÷b)^2 (a+ a+b ... Webb26 maj 2024 · Explanation: sin(A +B)sin(A− B) = sin2A −sin2B LHS = sin(A+ B)sin(A− B) Recall: sin(α − β) = sinαcosβ− cosαsinβ And sin(α +β) = sinαcosβ +cosαsinβ = (sinAcosB … Webb5 juni 2024 · = sin 2 A + sin(A -B) [sin(A –B) – 2 sinA cosB] We know that sin(A –B) = sinA cosB – cosA sinB = sin 2 A + sin(A -B) [sinA cosB – cosA sinB – 2 sinA cosB] = sin 2 A + … phill bascombe