Question

Sin3A+cos2B=2 :then cos2A+sin3B=?

Answer 1

Answer:

$\frac{1}{2}$

Step-by-step explanation:

First of all, from the basic rules of trigonometry, $sin\theta$ or $cos\theta$ have a maximum value of 1.

Therefore, in order for $sin3a+cos2b$ to be equals to 2, this means that both 'sin3a' and 'cos2b' have reached their maximum value of 1.

Hence,

$sin3a=1\:\:\:and\:\:\:cos2b=1\\3a=90^o\:\:\:\:and\:\:\:\:2b=0^o\\a=30^o\:\:\:\:\:and\:\:\:\:b=0^o$

Now, we need to evaluate:

$cos2a+sin3b\\=cos2(30^o)+sin3(0^o)\\=cos60^o+sin0^o\\=\frac{1}{2}$

Answer 2

Answer: the correct answer is 0.5=1/2

Step-by-step explanation:

Given, sin3a + cos2b = 2

sin3a=1 &cos2b=1

3a=90° & 2b=0°

a=30° & b=0°

Now we have to find

cos2a + sin3b

= cos60° + sin0°

=1/2=0.5