Cosb=-5/13 и sina>0 найти sin2b,sinb,tgb

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Given that cos b = -5/13 and sin a > 0, we can determine that sin b = √(1 - cos² b) = √(1 - (-5/13)²) = √(1 - 25/169) = √(144/169) = 12/13.

Now, we can find sin 2b by using the double angle identity sin 2b = 2sin b cos b.

sin 2b = 2(sin b)(cos b) = 2(12/13)(-5/13) = -120/169.

Finally, we can calculate tan b by using the identity tan b = sin b / cos b.

tgb = sin b / cos b = (12/13) / (-5/13) = -12/5.

Therefore, sin 2b = -120/169, sin b = 12/13, and tan b = -12/5.