 Hello and welcome to the session. In this session we will find the formula for trigonometric values of sum and difference of two angles. And we will evaluate the expressions like cos of alpha minus beta and sin of alpha minus beta where alpha and beta are angles. First of all we will derive the formula of cos of alpha minus beta and the formula of cos of alpha minus beta. Now let us find all nodes of a point on unit circle in terms of trigonometric ratios. Now consider a unit circle that is the circle with radius 1 and here o is the center of this circle. Now let p be any point on the circle or we can say let p be any point on the circle points of the circle. Now let us derive op and op is the radius of the circle which is equal to 1 unit. And now from the point p we will draw a perpendicular on x axis such that triangle opm is a right angle triangle and angle pom is equal to theta. Now let opm is equal to x and opm is equal to y. Then coordinates of the point p are given by the other pair xy. Now we need that in a right angle from angle the side opposite to the right angle is called the high continuous. Now if an angle is given to us as theta then the side opposite to this angle is called perpendicular and the remaining side is called the base. Now if we want to find sin theta from this triangle then sin theta will be equal to perpendicular upon high pattern use which is equal to ab upon ac and cos theta is equal to base upon high pattern use that is vc upon ac. And if here we have to find sin theta from this given triangle abc in which this is triangle theta then the side bc will be perpendicular side ab will be base and side ac will be high pattern use so sin theta will be equal to perpendicular upon high pattern use that is bc upon ac and cos theta is equal to base upon high pattern use that is ab upon ac. Similarly from this triangle that is the triangle opm which is the right angle triangle we can find sin theta here and the pom is theta so sin theta will be equal to perpendicular that is pm upon high pattern use that is op and this is equal to now pm is equal to y and op is the radius which is equal to 1 so sin theta is equal to y and cos theta is equal to base upon high pattern use which is equal to x upon 1 which is equal to x. So coordinates of p in terms of trigonometric ratios are given by the oddity pair cos theta sin theta because here x is equal to cos theta and y is equal to sin theta. So coordinates of p in terms of trigonometric ratios are cos theta sin theta. Thus in a unit circle that is the circle with radius 1 unit the oddity pair xy is given by the oddity pair cos theta sin theta and now we will derive the formula. Now let p and p dash be two primes on the circumference of this unit circle and here alpha and beta are the angles made by op and op dash respectively with x axis. So the coordinates of the point p will be cos alpha sin alpha and coordinates of the point p dash will be cos beta sin beta. Now let d be the distance between the points p and p dash. So by distance formula distance d will be equal to square beta as alpha minus cos beta whole square plus sin alpha minus sin beta whole square. Now square in both sides we get d square is equal to cos alpha minus cos beta whole square plus sin alpha minus sin beta whole square. Now using this identity we will open the brackets and this will be d square is equal to cos square alpha minus 2 cos alpha cos beta plus cos square beta and now let us open the second bracket and this will be sin square alpha minus 2 sin alpha sin beta plus sin square beta. Now we have a trigonometric identity that is cos square theta plus sin square theta is equal to 1. Now here also combining these terms we have cos square alpha plus sin square alpha and cos square beta plus sin square beta. That is we have d square is equal to cos square alpha plus sin square alpha plus cos square beta plus sin square beta minus 2 cos alpha cos beta minus 2 sin alpha sin beta. Now using this trigonometric identity cos square alpha plus sin square alpha is 1 and cos square beta plus sin square beta is again 1 and 1 plus 1 is 2. So this implies d square is equal to 2 minus 2 cos alpha cos beta minus 2 sin alpha sin beta. Now let us see equation number 1. Now earlier we have discussed law of the science that in a triangle A, B, C having sides with natures A, B, C and opposite angles capital A, capital B and capital C. And here we have A square is equal to B square plus C square minus 2 B sin into cos of capital A. Now in this diagram we have a triangle P over P dash with measure of sides P P dash is equal to d and OP and OP dash is equal to 1 unit. And always opposite to sides OP OP dash and P P dash are OP dash P OP P dash and P OP dash respectively. And from this diagram we can find the measure of angle P OP dash which is equal to angle P O X minus angle P dash O X this means angle P OP dash is equal to alpha minus beta. It means angle opposite to sides P P dash that is d is equal to alpha minus beta. Now using this law we have d square is equal to 1 square plus 1 square minus 2 into 1 into 1 into cos of alpha minus beta. And this implies d square is equal to 2 minus 2 cos of alpha minus beta. Now let this be equation number 2. Now this is the equation number 1 and this is equation number 2. So equating equation number 1 and 2 we get 2 minus 2 cos of alpha minus beta is equal to 2 minus 2 cos alpha cos beta minus 2 sin alpha sin beta. Further this implies minus 2 cos of alpha minus beta is equal to minus 2 cos alpha cos beta minus 2 sin alpha sin beta. Now dividing throughout by minus 2 this implies cos of alpha minus beta is equal to cos alpha cos beta plus sin alpha sin beta. So this is the required formula of cos of alpha minus beta and now let us find the formula of cos of alpha plus beta. For this we will replace beta by minus beta in this equation let it be equation number 3. And we have cos of alpha minus of minus beta which is equal to cos alpha cos of minus beta plus sin alpha sin of minus beta. Now we know that sin of minus theta is minus sin theta and cos of minus theta is cos theta. So this implies cos of alpha plus beta is equal to cos alpha and cos of minus beta will be cos beta plus sin alpha. And here sin of minus beta will be minus sin beta. So here we have cos of alpha plus beta is equal to cos alpha cos beta minus sin alpha sin beta. Now let us derive this formula. Now we can derive sin of alpha plus beta using cosine formula. Now we know that sin of theta is equal to cos of pi by 2 minus theta. Now let us put theta is equal to alpha plus beta here. This will be sin of alpha plus beta is equal to cos of pi by 2 minus of alpha plus beta as a whole. And this will be pi by 2 minus alpha minus beta. Now this can be written as cos pi by 2 minus alpha the whole minus beta. And now we will use this cosine formula for solving this further. Now let us apply this formula for cos of pi by 2 minus alpha the whole minus beta. And this is equal to now here in place of alpha we will put pi by 2 minus alpha. And this will be cos of pi by 2 minus alpha into cos beta plus sin of pi by 2 minus alpha into sin beta. Now we will use this formula and this is equal to now cos of pi by 2 minus alpha is sin alpha into cos beta plus sin of pi by 2 minus alpha will be cos alpha into sin beta. Therefore sin of alpha plus beta is equal to sin alpha cos beta plus cos alpha sin beta. Now let us find sin of alpha minus beta. Now replacing beta by minus beta in this equation that is equation 4 we get sin of alpha minus beta is equal to sin alpha plus beta minus cos alpha sin beta. Now let us derive the formula of tan of alpha plus beta and the formula of tan of alpha minus. Now we know that tan theta is equal to sin theta then cos theta. So here we can write tan of alpha plus beta is equal to sin of alpha plus beta upon cos of alpha plus beta. Now using this formula this is equal to sin alpha cos beta plus cos alpha sin beta whole upon cos alpha cos beta minus sin alpha sin beta. Now dividing numerator and denominator by the product cos alpha cos beta we get. Now here in the numerator we have sin alpha upon cos alpha plus sin beta upon cos beta whole upon 1 minus sin alpha. Sin alpha sin beta whole upon cos alpha cos beta. And this is equal to now sin alpha upon cos alpha is tan alpha plus sin beta upon cos beta is tan beta whole upon 1 minus tan alpha tan beta. So this is the required formula of tan of alpha plus beta. Similarly we can derive the formula of tan of alpha minus beta. So we have derived all these formulae. Now let us discuss an example. Here we have to find sin of 135 degrees using sum of angles formula. Now we know that 135 degrees is equal to 90 degrees plus 45 degrees. Therefore sin of 135 degrees is equal to sin of 90 degrees plus 45 degrees. So here we will use the formula of sin of alpha plus beta. So sin of 135 degrees which is written as sin of 90 degrees plus 45 degrees will be equal to now here alpha is 90 degrees and beta is 45 degrees. So using this formula this will be equal to sin of 90 degrees cos of 45 degrees plus cos of 90 degrees sin of 45 degrees. Now from the table of values sin 90 degrees is 1 into cos 45 degrees is 1 upon root 2 plus cos 90 degrees is 0 and sin 45 degrees is 1 upon root 2. So this is equal to 1 upon root 2 plus 0 which is equal to 1 upon root 2. So sin 135 degrees is equal to 1 upon root 2. So in this session we have derived all these formulae and this completes our session. Hope you all have enjoyed the session.