Question

$\huge\sf{Question}$

If A = 600 and B = 300, verify that

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B – sin A sin B

(iii) sin (A – B) = sin A cos B – cos A sin B

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B).

Don't Spam​

Answer 1

This will check conditions 2 and 3, even if 1 is already true. As your conditions can be quite expensive functions, you can get a good performance boost by using them.

There is one big caveat, NullReferences or similar problems. For example:

if(class != null && class.someVar < 20)

If class is null, the if-statement will stop after class != null is false. If you only use &, it will try to check class.someVar and you get a nice NullReferenceException. With the Or-Operator that may not be that much of a trap as it's unlikely that you trigger something bad, but it's something to keep in mind.

No one ever uses the single & or | operators though, unless you have a design where each condition is a function that HAS to be executed. Sounds like a design smell, but sometimes (rarely) it's a clean way to do stuff. The & operator does "run these 3 functions, and if one of them returns false, execute the else block", while the | does "only run the else block if none return false" - can be useful, but as said, often it's a design smell.

Answer 2

Answer:

Verified.

Step-by-step explanation:

600 can be written as 540° + 60°

                                   = 180°* 3 + 60°

                                   = 180° + 60°         (3 times 180 will lie on 180)

                                   = 60°                   ( multiple of π)

This will lie in the 3rd quadrant

In this question, we require sin 600° and cos 600°

Therefore sin (600°) = sin(180° + 60°)

In the third quadrant, sin is negative

Therefore, sin(180° + 60°) = - sin 60° = -√3/2

Similarly, cos (600°) = cos(180° + 60°)

In the third quadrant, cos is negative

Therefore, cos(180° + 60°) = - cos 60° = -1/2

Therefore, sin 600° = -√3/2 and cos 600° = -1/2 and tan 600° = √3

Similarly, sin 300° = -√3/2 and cos 300° = 1/2 and tan 300° = -√3

Similarly, sin 900° = 0 and cos 900° = -1 and tan 900° = 0

Here, directly substituting the value of A=600°, B = 300° and A+B = 900°

(i) LHS  = 0

   RHS = -√3/2 * 1/2 + -1/2 * -√3/2 = 0

(ii) LHS  = -1

   RHS = -1/2 * 1/2 - (-√3/2 * -√3/2) = -1

(iii) LHS  = -√3/2

   RHS = -√3/2 * 1/2 - (-1/2 * -√3/2) = -√3/2

(iii) LHS  = -√3

   RHS = [√3 - (-√3)]/ [ 1+ √3*(-√3)]= -√3

Hence verified.