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<\/figure>\n\n\n\nSum to Product Formulas Proof<\/h2>\n\n\n\n
Now, that we have discussed the sum to product formulas in trigonometry, let us derive these formulas using the product to sum formulas whose formulas are given by,<\/p>\n\n\n\n
- sin A cos B = (1\/2) [ sin (A + B) + sin (A – B) ] — (1)<\/li>
- cos A sin B = (1\/2) [ sin (A + B) – sin (A – B) ] — (2)<\/li>
- cos A cos B = (1\/2) [ cos (A + B) + cos (A – B) ] — (3)<\/li>
- sin A sin B = (1\/2) [ cos (A – B) – cos (A + B) ] — (4)<\/li><\/ul>\n\n\n\n
To derive the sum to product formulas, assume (p + q)\/2 = A and (p – q)\/2 = B. Now, taking the sum and difference of A and B, we have<\/p>\n\n\n\n
A + B = [(p + q)\/2] + [(p – q)\/2]<\/p>\n\n\n\n
= p\/2 + q\/2 + p\/2 – q\/2<\/p>\n\n\n\n
= p\/2 + p\/2<\/p>\n\n\n\n
= p<\/p>\n\n\n\n
A – B = [(p + q)\/2] – [(p – q)\/2]<\/p>\n\n\n\n
= p\/2 + q\/2 – p\/2 + q\/2<\/p>\n\n\n\n
= q\/2 + q\/2<\/p>\n\n\n\n
= q<\/p>\n\n\n\n
Now, substituting the values of A, B, A + B and A – B in the formulas (1), (2), (3), and (4), we have<\/p>\n\n\n\n
- sin [(p + q)\/2] cos [(p – q)\/2] = (1\/2) [ sin p + sin q ]
2 sin [(p + q)\/2] cos [(p – q)\/2] = sin p + sin q — (5)<\/li> - cos [(p + q)\/2] sin [(p – q)\/2] = (1\/2) [ sin p – sin q ]
2 cos [(p + q)\/2] sin [(p – q)\/2] = sin p – sin q — (6)<\/li> - cos [(p + q)\/2] cos [(p – q)\/2] = (1\/2) [ cos p + cos q ]
2 cos [(p + q)\/2] cos [(p – q)\/2] = cos p + cos q — (7)<\/li> - sin [(p + q)\/2] sin [(p – q)\/2] = (1\/2) [ cos q – cos p ]
-2 sin [(p + q)\/2] sin [(p – q)\/2] = cos p – cos q — (8)<\/li><\/ul>\n\n\n\nHence, we have derived the sum to product formulas which are given by the formulas (5), (6), (7) and (8).<\/p>\n\n\n\n
Using Sum to Product Formula<\/h2>\n\n\n\n
We use the sum to product formula to simplify and solve mathematical problems in trigonometry. In this section, we will understand how to apply the sum to product formulas with the help of solving a few examples.<\/p>\n\n\n\n
Example 1:<\/strong> Express the difference of cosines cos 4x – cos x as a product of trigonometric function using sum to product formulas.<\/p>\n\n\n\n
Solution:<\/strong> We know that cos A – cos B = -2 sin [(A + B)\/2] sin [(A – B)\/2]. Substituting A = 4x and B = x into this formula, we have<\/p>\n\n\n\n
cos 4x – cos x = -2 sin [(4x + x)\/2] sin [(4x – x)\/2]<\/p>\n\n\n\n
= -2 sin (5x\/2) sin (3x\/2)<\/p>\n\n\n\n
Answer:<\/strong> Hence, we can express the difference cos 4x – cos x as -2 sin (5x\/2) sin (3x\/2) as a product of trigonometric functions.<\/p>\n\n\n\n
Example 2:<\/strong> Evaluate the value of sin 15\u00b0 + sin 75\u00b0 using the sum to product formula.<\/p>\n\n\n\n
Solution:<\/strong> The formula required to find the value of sin 15\u00b0 + sin 75\u00b0 is sin A + sin B = 2 sin [(A + B)\/2] cos [(A – B)\/2]. Substituting A = 15\u00b0 and B = 75\u00b0 into the formula, we have<\/p>\n\n\n\n
sin 15\u00b0 + sin 75\u00b0 = 2 sin [(15\u00b0 + 75\u00b0)\/2] cos [(15\u00b0 – 75\u00b0)\/2]<\/p>\n\n\n\n
= 2 sin(90\u00b0\/2) cos (-60\u00b0\/2)<\/p>\n\n\n\n
= 2 sin 45\u00b0 cos 30\u00b0 — [Because cos(-x) = cos x]<\/p>\n\n\n\n
= 2 \u00d7 1\/\u221a2 \u00d7 \u221a3\/2 — [Because cos 30\u00b0 is equal to \u221a3\/2 and sin 45\u00b0 is equal to 1\/\u221a2]<\/p>\n\n\n\n
= \u221a3\/\u221a2<\/p>\n\n\n\n
= \u221a(3\/2)<\/p>\n\n\n\n
Answer:<\/strong> sin 15\u00b0 + sin 75 = \u221a(3\/2) using sum to product formula.<\/p>\n\n\n\n
Important Notes on Sum to Product Formula<\/strong><\/p>\n\n\n\n
- The sum to product formulas are used to express the sum and difference of trigonometric functions sines and cosines as products of sine and cosine functions.<\/li>
- We can derive the sum to product formula using the product to sum formulas in\u00a0trigonometry.<\/li>
- We can apply these formulas to simplify trigonometric problems.<\/li>
- The sum to product formula are:
- sin A + sin B = 2 sin [(A + B)\/2] cos [(A – B)\/2]<\/li>
- sin A – sin B = 2 sin [(A – B)\/2] cos [(A + B)\/2]<\/li>
- cos A – cos B = -2 sin [(A + B)\/2] sin [(A – B)\/2]<\/li>
- cos A + cos B = 2 cos [(A + B)\/2] cos [(A – B)\/2]<\/li><\/ul><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"
We can prove these sum to product formulas using the\u00a0product to sum formulas\u00a0in trigonometry. The sum to product formula are expressed as follows: sin A + sin B = 2 sin [(A + B)\/2] cos [(A – B)\/2] sin A – sin B = 2 sin [(A – B)\/2] cos [(A + B)\/2] cos A […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[229],"tags":[],"_links":{"self":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7814"}],"collection":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/comments?post=7814"}],"version-history":[{"count":0,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/posts\/7814\/revisions"}],"wp:attachment":[{"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/media?parent=7814"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/categories?post=7814"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mudassirbackup.infinitycodestudio.com\/index.php\/wp-json\/wp\/v2\/tags?post=7814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- sin [(p + q)\/2] cos [(p – q)\/2] = (1\/2) [ sin p + sin q ]