Sin a-b formula

Solution : The formula of sin (A + B + C) is sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C. Proof : We have, sin (A + B + C) = sin ((A + B) + C) = sin (A + B) cos C + cos (A + B) sin C sin (A + B + C) = (sin A cos B + cos A sin B) cos C + (cos A cos B – sin A sin B) sin C sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C Post navigation

Sin(a + b) Sin(a + b) is one of the important trigonometric identities used in trigonometry. It is one of sum and difference formulas. It says sin (a + b) = sin a cos b + cos a sin b. We use the sin(a + b) identity to find the value of the sine trigonometric function for the sum of angles. The expansion of sin a plus b formula helps in representing the sine of a compound angle in terms of sine and cosine trigonometric functions. Let us understand the sin(a+b) identity and its proof in detail in the following sections. 1. 2. 3. 4. Proof of Sin(a + b) Formula The proof of expansion of sin(a + b) formula can be done geometrically. Let us see the stepwise derivation of the formula for the To prove: sin (a + b) = sin a cos b + cos a sin b Construction: Assume a rotating line OX and let us rotate it about O in the anti-clockwise direction. OX makes out an acute ∠XOY = a, from starting position to its initial position. Again, the rotating line rotates further in the same direction and starting from the position OY, thus making out an acute angle given as, ∠YOZ = b. ∠XOZ = a + b < 90°. On the bounding line of the compound angle (a + b) take a point P on OZ, and draw PQ and PR perpendiculars to OX and OY respectively. Again, from R draw perpendiculars RS and RT upon OX and PQ respectively. Proof: From triangle PTR we get, ∠TPR = 90° - ∠PRT = ∠TRO = alternate ∠ROX = a. Now, from the right-angled triangle PQO we get, sin (a + b) = PQ/OP = (PT + TQ)/OP = PT/OP + TQ/OP = PT/OP + RS/OP ...

The standard formula for #sin(A+B)# is: #sin(A+B) = sin(A)cos(B)+cos(A)sin(B)# Now #sin(-B) = -sin(B)# and #cos(-B) = cos(B)#, so #sin(A-B) = sin(A)cos(B)-cos(A)sin(B)# So: #sin(A+B)*sin(A-B)# #= (sin A cos B + cos A sin B)(sin A cos B - cos A sin B)# #= (sin A cos B)^2 - (cos A sin B)^2# ...using the identity #(p+q)(p-q) = p^2-q^2#. #= sin^2Acos^2B-sin^2Bcos^2A# #= sin^2A(1-sin^2B)-sin^2B(1-sin^2A)# ...using #sin^2 theta + cos^2 theta = 1# (Pythagoras) #= sin^2A-sin^2B-sin^2Asin^2B+sin^2Bsin^2A# #= sin^2A - sin^2B#

Sin(a + b) Sin(a + b) is one of the important trigonometric identities used in trigonometry. It is one of sum and difference formulas. It says sin (a + b) = sin a cos b + cos a sin b. We use the sin(a + b) identity to find the value of the sine trigonometric function for the sum of angles. The expansion of sin a plus b formula helps in representing the sine of a compound angle in terms of sine and cosine trigonometric functions. Let us understand the sin(a+b) identity and its proof in detail in the following sections. 1. 2. 3. 4. Proof of Sin(a + b) Formula The proof of expansion of sin(a + b) formula can be done geometrically. Let us see the stepwise derivation of the formula for the To prove: sin (a + b) = sin a cos b + cos a sin b Construction: Assume a rotating line OX and let us rotate it about O in the anti-clockwise direction. OX makes out an acute ∠XOY = a, from starting position to its initial position. Again, the rotating line rotates further in the same direction and starting from the position OY, thus making out an acute angle given as, ∠YOZ = b. ∠XOZ = a + b < 90°. On the bounding line of the compound angle (a + b) take a point P on OZ, and draw PQ and PR perpendiculars to OX and OY respectively. Again, from R draw perpendiculars RS and RT upon OX and PQ respectively. Proof: From triangle PTR we get, ∠TPR = 90° - ∠PRT = ∠TRO = alternate ∠ROX = a. Now, from the right-angled triangle PQO we get, sin (a + b) = PQ/OP = (PT + TQ)/OP = PT/OP + TQ/OP = PT/OP + RS/OP ...

Solution : The formula of sin (A + B + C) is sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C. Proof : We have, sin (A + B + C) = sin ((A + B) + C) = sin (A + B) cos C + cos (A + B) sin C sin (A + B + C) = (sin A cos B + cos A sin B) cos C + (cos A cos B – sin A sin B) sin C sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C Post navigation

The standard formula for #sin(A+B)# is: #sin(A+B) = sin(A)cos(B)+cos(A)sin(B)# Now #sin(-B) = -sin(B)# and #cos(-B) = cos(B)#, so #sin(A-B) = sin(A)cos(B)-cos(A)sin(B)# So: #sin(A+B)*sin(A-B)# #= (sin A cos B + cos A sin B)(sin A cos B - cos A sin B)# #= (sin A cos B)^2 - (cos A sin B)^2# ...using the identity #(p+q)(p-q) = p^2-q^2#. #= sin^2Acos^2B-sin^2Bcos^2A# #= sin^2A(1-sin^2B)-sin^2B(1-sin^2A)# ...using #sin^2 theta + cos^2 theta = 1# (Pythagoras) #= sin^2A-sin^2B-sin^2Asin^2B+sin^2Bsin^2A# #= sin^2A - sin^2B#