Prove that sinx+cosx=sin2x+cos2x?

Question

1) Prove that sinx +cosx=sin2x+cos2x
2) How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 ,if the repetition of the digits are not allowed?

Answer 1

1. Not true. As others above have shown

2.
one digit: 3, 5, 7
two digits: 35, 37, 53, 57, 73, 75
three digits: 305, 307, 357, 375, 503, 507, 537, 573, 703, 705, 735, 753.

There are 21 numbers.

Answer 2

(1)
If it's true, it has to be true for any x... right?
Ok, I choose x = π/4
Sin(π/2) + cos(π/2) = 1 + 0 = 1
Sin(π/4) + cos(π/4) = 2√(2/4) = √2 ≠1
Untrue... Proof by counter example.

(2)
0 is not prime.
3,5,7... that's it.
53 is, for example, prime, but the digits 3 and 5 have both been used.
Ok, I was moved to write an algorithm...
3, 5, 7, 37, 53, 73, 307, 503... all the rest either don't contain the numbers, aren't prime, or have duplicated digits.

Somebody doesn't seem to think my answer is correct. I'm presuming it's the numbers problem... Here's the results...

Note, if a number isn't prime, a set of factors is provided... there may be others, but only one set is required to prove a number's not prime.
3 . . . PRIME
5 . . . PRIME
7 . . . PRIME
30 . . . 2 x 15
33 . . . 3 x 11
35 . . . 5 x 7
37 . . . PRIME
50 . . . 2 x 25
53 . . . PRIME
55 . . . 5 x 11
57 . . . 3 x 19
70 . . . 2 x 35
73 . . . PRIME
75 . . . 3 x 25
77. . . 7 x 11
300 . . 2 x 150
303 . . 3 x 101
305 . . 5 x 61
307 . . PRIME
330 . . 2 x 165
333 . . 3 x 111
335 . . 5 x 67
337 . . Dup 3s
350 . . 2 x 175
353 . . Dup 3s
355 . . 5 x 71
357 . . 3 x 119
370 . . 2 x 185
373 . . Dup 3s
375 . . 3 x 125
377 . . 13 x 29
500 . . 2 x 250
503 . . PRIME
505 . . 5 x 101
507 . . 3 x 169
530 . . 2 x 265
533 . . 13 x 41
535 . . 5 x 107
537 . . 3 x 179
550 . . 2 x 275
553 . . 7 x 79
555 . . 3 x 185
557 . . Dup 5s
570 . . 2 x 285
573 . . 3 x 191
575 . . 5 x 115
577 . . Dup 7s
700 . . 2 x 350
703 . . 19 x 37
705 . . 3 x 235
707 . . 7 x 101
730 . . 2 x 365
733 . . Dup 3s
735 . . 3 x 245
737 . . 11 x 67
750 . . 2 x 375
753 . . 3 x 251
755 . . 5 x 151
757 . . Dup 7s
770 . . 2 x 385
773 . . Dup 7s
775 . . 5 x 155
777 . . 3 x 259

Sorry if you don't like it... it's true.

Answer 3

1) It is not right.
Counter example: x = pi/4
LHS = sqrt(2)
RHS = 1

2)
1 digit numbers: 3C1 = 3, pick one of 3, 5, 7
2 digit numbers: 3*2 = 6, since you can't pick 0
3 digit numbers: 3! + 3*2 = 12, 3! means without 0, and 3*2 means with 0 in the middle
4 digit numbers: 0, since any 4 digit number is above the limit of 1000.
Total = 21

Answer 4

2) sice the number is odd therefore number of ways to fill the units place is 3(3,5,7). Now the ways to fill the tens place is again 3(one number which has been used in the units place and one '0'). Now the ways th fill the hundreds place is 2(2 no.s have been used in the other 2 places)
Therefore the no. of odd numbers that can be formed=3x3x2=18

Answer 5

1. You cannot prove this because it is not true!

2. If you don't have the time to work this out neither do I.